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9th Maths - Quadrilateral and Its Types

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The document provides information about quadrilaterals including definitions, types of quadrilaterals, and properties of quadrilaterals. It begins by defining a quadrilateral and listing the eight main types: parallelogram, rectangle, rhombus, square, trapezium, isosceles trapezium, kite I, and kite II. It then presents eight theorems proving various properties of quadrilaterals, such as the sum of interior angles equaling 360°, opposite sides being congruent, and diagonals bisecting each other. It concludes by providing three example problems solving for angles and side lengths of various quadrilaterals. The overall document serves as a comprehensive overview of quadril

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9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counseling 1
9011041155 / 9011031155 Quadrilaterals We know that quadrilateral is a closed figure with four sides. But by definition suppose P, Q, R and S are four points in a plane such that i. 3 of these points are not collinear ii. If sides PQ, QR, RS, SP are such that any of the two such segments have a common point, it is an endpoint only. iii. If we draw a line containing any one of these segments the remaining two points lie on the same side of this line and iv. If seg PR and seg QS intersect in the points other than P, Q, R, S then the union of 4 segments PQ, QR, RS, SP is called a quadrilateral. is a symbol of ‘quadrilateral’. 2
9011041155 / 9011031155 (And Interior of the quadrilateral is a convex set but quadrilateral is not a convex set) Each quadrilateral has 4 sides, 4 angles, 4 vertices, 2 diagonals, 4 pairs of adjacent angles, 2 pairs of opposite sides, 4 pairs of adjacent or consecutive sides and 2 pairs of opposite angles. There are main 8 types of quadrilaterals (as shown in the figures they are i. Parallelogram ii. Rectangle iii. Rhombus iv. Square v. Trapezium (non parallel sides are not congruent) vi. Isosceles trapezium vii. Kite viii. Kite (with all side, congruent) 3
9011041155 / 9011031155 Table S. No. Quadrilateral Type 1. Parallelogram 2. Rectangle 3. Rhombus 4. Square 4
9011041155 / 9011031155 5. Trapezium 6. Isosceles Trapezium 7. Kite I 8. Kite II 5
9011041155 / 9011031155 Property of a quadrilateral Theorem 1 [Angle sum property of a quadrilateral] The sum of the measures of all angles of a quadrilateral is 360º Given : To prove : Construction : ABCD is a given quadrilateral ∠A + ∠B + ∠C + ∠D = 360º Join the points A and C 6
9011041155 / 9011031155 Proof : In ABC, ∠BAC + ∠ ABC + ∠BCA = 180º .....sum of angles of ABC .... (1) and, In ADC, ∠DAC + ∠ADC + ∠DCA = 180º .....sum of angles of ADC .... (2) 7
9011041155 / 9011031155 Adding (1) and (2) we get ∠BAC + ∠ABC + ∠BCA + ∠DAC + ∠ADC + ∠DCA = 180º + 180º ∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA + ∠DCA + ∠ADC = 360º ......(3) But ∠BAC + ∠DAC = ∠BAD ....Angle addition property....(4) and ∠BCA + ∠DCA = ∠BCD ....Angle addition property....(5) ∴ from (3), (4) and (5) we get 8
9011041155 / 9011031155 ∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º i.e. ∠A + ∠B + ∠C + ∠D = 360º Hence, sum of the measures of all angles of a quadrilateral is 360º 9
9011041155 / 9011031155 Theorem 2 The opposite sides of a parallelogram are congruent. Given ABCD is a parallelogram in which : side AB || side CD and side BC || side DA : Construction : Proof : side AB side CD and side BC To prove side DA Draw diagonal AC. ABCD is a parallelogram in which side AB || side CD and seg AC is a transversals. 10
9011041155 / 9011031155 ∴∠BAC ∠DCA ...(alternate angle)....(1) Similarly, side BC || side DA and seg AC is a transversal. ∴ ∠BCA ∠DAC ......(alternate angle) ...(2) In ABC and ∠BAC seg AC ∠BCA ∴ CDA ∠DCA ......(from 1) seg CA .......(common side) ∠ DAC ABC CDA 11 ........(from 2) .......(ASA test)
9011041155 / 9011031155 ∴ side AB side CD and side BC ......(c.s.c.t) side DA ......(c.s.c.t.) Hence, the opposite sides of a parallelogram are congruent. 12
9011041155 / 9011031155 Theorem 3 If opposite sides of a quadrilateral are congruent then the quadrilateral is parallelogram. Given : In ABCD side AB side DC and side AD To prove : Construction : side BC ABCD is a parallelogram Draw a diagonal BD 13
9011041155 / 9011031155 Proof : In ABD and CDB side AB side DC side AD side BC side BD side DB ......(given) ....(common side) ∴ ABD CDB ∴ ∠ABD ∠CDB .....(SSS test) ......(c.a.c.t.) ∴ side AB || side CD .......(1) ...(alternate angles test) 14
9011041155 / 9011031155 Similarly, we can prove that side AD || side BC ......(2) ....(alternate angels test) ∴ from (1) and (2) we have ABCD is a parallelogram Hence, if opposite sides of a quadrilateral are congruent then it is a parallelogram. 15
9011041155 / 9011031155 Theorem 4 The opposite angles of a parallelogram are congruent Given PQRS is a parallelogram : To prove : ∠SPQ ∠QRS and ∠PSR Construction : Proof : ∠RQP Draw a diagonal SQ PQRS is a parallelogram ∴ side PS || side QR and seg SQ is a transversal ∴ ∠PSQ ∠RQS ..(Alternate angles) ...(1) 16
9011041155 / 9011031155 Also, side PQ || side SR and seg SQ is a transversal. ∴ ∠PQS ∠RSQ ...(Alternate angles)..(2) In PQS and ∠PSQ RSQ ∠RQS ......from (1) side SQ side QS ....(common side) ∠PQS ∴ ∠RSQ PQS ......from (2) RSQ ......(ASA test) ∴ ∠SPQ 17 ∠QRS ......(c.a.c.t.)
9011041155 / 9011031155 Similarly, we can prove by drawing diagonal PR. ∠PSR ∠ RQP Hence, the opposite angles of a parallelogram are congruent. 18
9011041155 / 9011031155 Theorem 5 A quadrilateral is a parallelogram if its opposite angles are congruent. Given : PQRS is a parallelogram in which ∠SPQ ∠PQR To prove : Proof : ∠QRS and ∠RSP PQRS is a parallelogram Let ∠SPQ = ∠QRS = xº And ∠PQR = ∠RSP = yº ......Opposite angle of a quadrilateral. 19
9011041155 / 9011031155 ∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º. …(Angle sum property of a quadrilateral) ∴ x + y + x + y = 360º ∴ 2x + 2y = 360º ∴ x + y = 180º .....(dividing by 2) ∴ ∠SPQ + ∠RSP = 180º ∴ side PQ || side SR ... (interior angles test) ....(1) 20
9011041155 / 9011031155 Similarly, we can prove that side PS || side QR .......(2) ∴ PQRS is a parallelogram ...... from (1) and (2) Hence, if opposite angles of a quadrilateral are congruent, then it is a parallelogram. 21
9011041155 / 9011031155 Theorem 6 The diagonals of a parallelogram bisect each other. Given : ABCD is a parallelogram in which the diagonals AC and BD intersect in M. To prove : seg AM seg CM and seg BM Proof : since seg DM ABCD is a parallelogram side AB || side CD and AC is a transversal. ∴ ∠BAC ∠DCA ...(Alternate angles) i.e. ∠BAM ∠DCM ...(A – M – C)..(1) 22
9011041155 / 9011031155 Also, side AB || side DC and seg DB is a transversal. ∴ ∠ABD ∠CDB ...(Alternate angles) i.e. ∠ABM ∠CDM ...(B – M – D) ..(2) Now, In ABM and ∠BAM ∠DCM ......(from 1) side AB ∠ABM ∴ CDM side DC .....(opposite sides) ∠CDM ......(from 2) ABM CDM ......(ASA test) 23
9011041155 / 9011031155 ∴ seg AM seg CM and seg BM .....(c.s.c.t.) seg DM Hence, diagonals of a parallelogram bisect each other. 24
9011041155 / 9011031155 Theorem 7 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Given : PQRS is a quadrilateral in which diagonals PR and QS intersect in M. seg PM seg QM To prove : Proof : seg RM and seg SM PQRS is a parallelogram In PMQ and seg PM RMS seg RM 25 ......(given)
9011041155 / 9011031155 ∠PMQ ∠RMS .....(verticallyopposite angles) seg QM ∴ seg SM .....(given) PMQ RMS ∴ ∠PQM ∠RSM i.e. ∠PQS ......(SAS test) ......(c.a.c.t) ∠RSQ ......(S – M – Q) ∴ side PQ || side SR ....(alternate angles test) ...(1) Similarly, we can prove that side PS || side QR 26 .....(2)
9011041155 / 9011031155 PQRS is a parallelogram .........from (1) and (2) Hence, if the diagonals of a quadrilateral bisect each other then it is a parallelogram. 27
9011041155 / 9011031155 Theorem 8 A quadrilateral is a parallelogram if a pair of opposite sides is parallel and congruent. Given LMNK is a given quadrilateral in : which side LM || side NK and side LM To prove : Construction : side NK LMNK is a parallelogram Draw diagonal MK 28
9011041155 / 9011031155 Proof : since LMNK is a quadrilateral in which side LM || side NK and seg MK is a transversal. ∴∠LMK ∠NKM ..(Alternate angles) ...(1) Now, In KLM and MNK seg LM seg NK .....(given) ∠LMK ∠NKM seg KM .......(from 1) seg MK .......(common side) ∴ KLM MNK ......(SAS test) 29
9011041155 / 9011031155 ∴ ∠LKM ∠NMK .......(c.a.c.t) ∴ side LK || side MN .....(alternate angles test) ....(2) and side LM || side NK ...(given) .....(3) ∴ from (2) and (3) we have LKMN is a parallelogram Hence, if a pair of opposite sides of a quadrilateral is parallel and congruent then the quadrilateral is a parallelogram. 30
9011041155 / 9011031155 Sums 1. The angles of a quadrilateral are in the ratio 2 : 3 : 5 : 8. Find all the angles of the quadrilateral. Sol. The angles of the quadrilateral are in ratio 2 : 3 : 5 : 8. Let the measures of the four angles are 2xº, 3xº, 5xº and 8xº. According to the angle sum property of a quadrilateral, we have 2xº + 3xº + 5xº + 8xº = 360º ∴ 18xº = 360º 3600 ∴x = 18 0 ∴ x 0 = 200 2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º, 5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º ∴ The measures of the angles of the quadrilateral are 40º, 60º, 100º and 160º. 31
9011041155 / 9011031155 2. In the figure, PQRS is a quadrilateral. The bisectors of ∠P and ∠Q meet at the point A. ∠S = 50º, ∠R = 110º. Find the measure of∠PAQ. Sol. Refer above fig. According to the angle sum property of a quadrilateral, ∠SPQ + ∠PQR + ∠R + ∠S = 360º ∴ ∠SPQ + ∠PQR + 110º + 50º = 360º … (substituting the given values) 32
9011041155 / 9011031155 ∴ ∠SPQ + ∠PQR + 160º = 360º ∴ ∠SPQ + ∠PQR = 360º - 160º ∴ ∠SPQ + ∠PQR = 200º … (1) Now, it is given that Ray PA and ray QA are the bisectors of ∠SPQ and ∠RQP respectively. ∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP ∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2) From (2) and (1), ∠APQ + ∠AQP = ½ (200º) ∴ ∠APQ + ∠AQP = 100º 33 … (3)
9011041155 / 9011031155 Also, In APQ, ∠PAQ + ∠APQ + ∠AQP = 180º … (Angles of a triangle) ∴ ∠PAQ + 100º = 180º … [From (3)] ∴ ∠PAQ = 180º - 100º ∴ ∠PAQ = 80º ∴The measure of ∠PAQ is 80º. 34
9011041155 / 9011031155 3. The sides BA and DC of quadrilateral ABCD are extended as shown in the figure. Prove that m + n = p + q. Proof : Consider ∠DAB b = a and ∠BCD = c. According to the angle sum property of a quadrilateral, In quadrilateral ABCD, a + q + c + p = 360º 35 … (1)
9011041155 / 9011031155 From (2), m + c + n + a = 360º … (3) From (1) and (3), a+q+c+p=m+c+n+a ∴q+p=m+n i.e. m + n = p + q. 36
9011041155 / 9011031155 4. In a parallelogram ABCD, ∠A = xº, ∠B = (3x + 20)º. Find x, ∠C and ∠D. Sol. We know that the adjacent angles of a parallelogram are supplementary. ∴ ∠A + ∠B = 180º Put the given values ∴ xº + 3xº + 20º = 180º ∴ 4xº = 180º - 20º ∴ 4xº = 160º 37
9011041155 / 9011031155 ∴ xº = 40º 3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º ∴ ∠A = 40º and ∠B = 140º Also, the opposite angles of a parallelogram are congruent. ∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º ∴ x = 40. ∠ C = 40º, ∠ D = 140º. 38
9011041155 / 9011031155 5. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the lengths of all the sides of the parallelogram. Sol: Consider that the smaller of the two adjacent sides be x cm. Then, it is given that the other side is (x + 25) cm Now, the perimeter of a parallelogram is equal to the sum of the lengths of four sides. ∴ x + (x + 25) + x + (x + 25) = 150 ∴ 4x + 50 = 150 ∴ x = 25 and x + 25 = 25 + 25 = 50 ∴ The lengths of the sides of the parallelogram ∴ 4x = 150 – 50 ∴ 4x = 100 are 25 cm, 50 cm, 25 cm and 50 cm. 39
9011041155 / 9011031155 6. In the figure, WXYZ is a parallelogram. From the information given in the figure, find the values of x and y. Sol. As XY & ZW are opposite sides of a parallelogram side XY || side ZW seg XZ is a transversal. ∴ ∠ ZXY = ∠ XZW ∴ 4y = 28º … (alternate angles) … (given) ∴ y = 7º 40
9011041155 / 9011031155 Similarly, side XW || side YZ and seg XZ is the transversal. ∴ ∠ ZXW = ∠ XZY. ∴ 10x = 60º … (given) ∴ x = 6º. ∴The value of x is 6º and that of y is 7º. 41
9011041155 / 9011031155 7. Prove that two opposite vertices of a parallelogram are equidistant from the diagonal not containing these vertices. Given : 1. ABCD is a parallelogram. 2. Seg AM ⊥ diagonal BD and Seg CN ⊥ diagonal BD. We have to prove seg AM ≅ seg CN. First, draw diagonal AC intersecting diagonal BD in the point P. 42
9011041155 / 9011031155 Proof : P is the point of intersection of the diagonals of parallelogram ABCD. ∴ seg AP ≅ seg CP … (1) In APM and CPN, seg AP ≅ seg CP … [From (1)] ∠ APM ≅ ∠ CPN … (vertically opposite angles) ∠ AMP ≅ ∠CNP … (each a right angle) 43
9011041155 / 9011031155 ∴ APM ≅ CPN … (SAA test) ∴ seg AM ≅ seg CN … (c.s.c.t.) Therefore, the opposite vertices of a parallelogram are at equal distance from the diagonal not containing these vertices. 44
9011041155 / 9011031155 8. The ratio of two sides of a parallelogram is 3 : 4. If its perimeter is 112 cm, find the lengths of the sides of the parallelogram. Sol: It is given that the ratio of the lengths of two sides of the parallelogram is 3:4. Let the lengths of those two sides be 3x cm and 4x cm respectively. We know that the opposite sides of a parallelogram are congruent. ∴ The sides of the parallelogram are 3x cm, 4x cm, 3x cm and 4x cm. Now, the perimeter of a parallelogram = the sum of the lengths of four sides ∴ 112 = 3x + 4x + 3x + 4x … (given perimeter 112 cm) 45
9011041155 / 9011031155 ∴ 14x = 112 ∴ x=8 3x = 3 × 8 = 24 and 4x = 4 × 8 = 32 ∴ The lengths of the sides of the given parallelogram are 24 cm, 32 cm, 24 cm and 32 cm. 46
9011041155 / 9011031155 Definitions Parallelogram A quadrilateral is a parallelogram if its opposite sides are parallel. Rectangle A parallelogram (or a quadrilateral) in which each angle is a right angle, is called a rectangle. Rhombus A quadrilateral having all sides congruent is called a rhombus. Square A quadrilateral is called a square if all its sides and angles are congruent. 47
9011041155 / 9011031155 Trapezium If only one pair of opposite sides is parallel, the quadrilateral is said to be a trapezium. Isosceles trapezium A trapezium in which nonparallel sides are congruent, it is called as isosceles trapezium. Kite If in ABCD, AB AD and CB CD and diagonal AC is perpendicular bisector of diagonal BD, then ABCD is called as a kite. 48
9011041155 / 9011031155 Theorem 9 Diagonals of a rectangle are congruent. Theorem 10 If diagonals of a parallelogram are congruent then it is a rectangle. 49
9011041155 / 9011031155 Theorem 11 Diagonals of a rhombus are perpendicular bisectors of each other. Theorem 12 If the diagonals of a quadrilateral bisect each other at right angle then the quadrilateral is a rhombus. 50
9011041155 / 9011031155 Theorem 13 Diagonals of a square are congruent and perpendicular bisectors of each other. Theorem 14 If diagonals of quadrilateral are congruent and perpendicular bisectors of each other then it is square. 51
9011041155 / 9011031155 Sums 1. Adjacent sides of a rectangle are of lengths 7 cm and 24 cm. Find the lengths of its diagonals. Sol. Consider ABCD as a rectangle in which AB = 24 cm and BC = 7 cm. In right-angled ABC, By Pythagoras’ theorem, AC² = AB² + BC² = (24)² + (7)² = 576 + 49 = 625 ∴ AC = 25 52
9011041155 / 9011031155 But, the diagonals of a rectangle are congruent. ∴ BD = AC = 25 ∴ The length of diagonal AC = 25 cm and that of BD = 25 cm. 53
9011041155 / 9011031155 2. ABCD is a trapezium in which AB || DC. M and N are the midpoints of side AD and side BC respectively. If AB = 12 cm and MN = 14 cm, find CD. Sol. The length of the segment joining the midpoints of non - parallel sides of a trapezium is half the sum of the lengths of its parallel sides. ∴ MN = ½ (AB + CD) ∴ 14 = ½ (12 + CD) … (substituting the given values) 54
9011041155 / 9011031155 ∴ 28 = 12 + CD … (Multiplying both the sides by 2) ∴ CD = 28 – 12 ∴ CD = 16. 55
9011041155 / 9011031155 3. Prove that the base angles of an isosceles trapezium are congruent. Given : □ ABCD is an isosceles trapezium in which side AB || side CD and side AD To prove : Construction : ∠ADC side BC. ∠BCD. Draw seg BE || side AD and D-E-C. 56
9011041155 / 9011031155 Proof : IN □ABED, side AB || side DE. … (given) seg BE || side AD … (construction ) ∴ □ABED is a parallelogram. The opposite sides and the opposite angles of a parallelogram are congruent. ∴ side AD and ∠D 57 side BE … (1) ∠ABE … (2)
9011041155 / 9011031155 side AD ≅ side BC … (given) … (3) From (1) and (3), side BE ≅ side BC ∴ ∠ BEC ≅ ∠C …(Isosceles triangle theorem) … (4) side AB || side DC … (given) side BE is the transversal. ∴ ∠ABE ≅ ∠BEC … (alternate angles) … (5) From (2), (4) and (5), ∠D ≅ ∠C. i.e. ∠ADC ≅ ∠BCD. [Similarly, we can prove that ∠DAB 58 ∠CBA.]
9011041155 / 9011031155 4. The lengths of the diagonals PR and QS of a rhombus are 20 cm and 48 cm respectively. Find the length of each side of the rhombus. Sol. We know that the diagonals of a rhombus bisect each other at right angles. ∴ MQ = ½ QS = ½ × 48 cm ∴ MQ = 24 cm MP = ½ PR = ½ × 20 cm ∴ MP = 10 cm. ∠ PMQ = 90º 59
9011041155 / 9011031155 In right-angled PMQ, By Pythagoras’ theorem PQ² = PM² + QM² = (10)² + (24)² = 100 + 576 = 676 ∴ PQ = 26. But all the sides of a rhombus are congruent. ∴ The length of each side of the given rhombus is 26 cm. 60
9011041155 / 9011031155 5. ABCD is a kite in which diagonal AC and diagonal BD intersect at point O. ∠ OBC = 20º and ∠OCD = 40º. Find : i. ∠ABC ii. ∠ADC iii. ∠BAD. Sol. Refer above fig, The diagonal joining the unequal sides of a kite bisects the other diagonal at right angles. ∴ Each angle at O is a right angle. Also, side BA side BC. 61
9011041155 / 9011031155 Side DA side DC and seg AO seg CO … (perpendicular bisector theorem) … (1) In BOC, ∠BOC = 90º, ∠OBC = 20º … (given) ∴ ∠OCB = 180º - (90º + 20º) = 180º - 110º ∴ ∠OCB = 70º … (2) 62
9011041155 / 9011031155 In BAC, BA = BC … [From (1)] ∴ ∠BCA = ∠BAC … (angles opposite to equal sides) i.e. ∠OCB = ∠OAB ∴ ∠OAB = 70º … [From (2)] In DAC, DA = DC ∴ ∠DCA = ∠DAC … [From (1)] … (angles opposite to equal sides) i.e. ∠DCO = ∠DAO 63 … (3)
9011041155 / 9011031155 ∴ ∠DAO = 40º … [Given : ∠DCO = 40º] … (4) From (3) and (4), ∠OAB + ∠DAO = 70º + 40º = 110º … (5) ∠BAO + ∠DAO = ∠BAD … (Angle addition postulate) … (6) From (5) and (6), ∠BAD = 110º. In DAC, ∠ADC + ∠DAC + ∠DCA = 180º … (angles of a triangle) ∴ ∠ ADC + 40º + 40º = 180º … [From (4) and given] 64
9011041155 / 9011031155 ∴ ∠ADC = 180º - 80º ∴ ∠ADC = 100º … (7) In BAC, ∠ABC + ∠BAC + ∠BCA = 180º … (angles of a triangle) ∴ ∠ABC + 70º + 70º = 180º … [From (2) and 3)] ∴ ∠ABC = 180º - 140º ∴ ∠ABC = 40º Ans. : i. ∠ABC = 40º ii. ∠ADC = 100º iii. ∠BAD = 110º. 65
9011041155 / 9011031155 Theorem 15 Diagonals of an isosceles trapezium are congruent. Given ABCD is an isosceles trapezium in : which seg AC and seg BD are diagonals side AD side BC and side AB || side DC To Prove : Construction : seg AC seg BD Draw seg AM ⊥ seg DC, seg BN ⊥ seg DC Proof : In ∆AMD and ∆BNC side AD side BC 66 ….(given)
9011041155 / 9011031155 ∠AMD seg AM ∠BNC …..(90º each) seg BN …..(perpendicular distance between two parallel lines) ∴ ∆AMD ∆BNC ……(hypotenuse side test) ∴ ∠ADM ∠BCN …..(c. a. c. t.) i.e. ∠ADC ∠BCD .(D - M - C)…(1) Now, In ∆ADC and ∆BCD side AD ∠ADC side BC ∠BCD 67 ….(given) …..(from (1))
9011041155 / 9011031155 side DC ∴ ∆ADC side CD …..(common side) ∆BCD ∴ seg AC …..(SAS test) seg BD ……(c.s.c.t.) Hence, diagonals of an isosceles trapezium are congruent. 68
9011041155 / 9011031155 Theorem 16 (Intercepts made, by three parallel lines) If three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. Given : line || line m || line n and lines t1 and t2 are transversals. seg AB Prove : Construction : seg BC seg DE seg EF Draw a line GI parallel to the line t1 through E. 69
9011041155 / 9011031155 Proof : seg AG || seg BE ……(given) seg AB || seg GE ……. (construction) ∴ ABEG is a parallelogram. ∴ seg AB seg GE…(opposite sides of a parallelogram …(1) Similarly, ∴ seg BC BCIE is a parallelogram. seg EI …..(opposite sides of a parallelogram….(2) But seg AB seg BC ….. (given)...(3) Hence seg GE seg EI ….(4) from (1), (2) and (3) 70
9011041155 / 9011031155 Now, In ∆GED and ∆IEF ∠DGE ∠FIE …(alternate angles) seg GE ∠GED seg EI …..(from (4)) ∠IEF ……(vertically opposite angles) ∴ ∆GED ∴ seg DE ∆IEF ……(ASA test) seg EF ……(c.s.c.t.) Hence, if three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. 71
9011041155 / 9011031155 Theorem 17 Mid - Point theorem The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it. Given : In ∆ABC, P and Q are midpoints of sides AB and AC respectively. To prove : i. seg PQ || side BC ii. seg PQ = 1 side BC 2 Construction : Take the point R on ray PQ such that seg PQ seg QR. Join C and R 72
9011041155 / 9011031155 Proof : In ∆AQP and ∆CQR seg AQ ∠AQP seg CQ ……(given) ∠CQR ….(vertically opposite angles) seg PQ seg RQ ……(construction) ∴ ∆AQP ∆CQR …… (SAS test) seg AP seg CR ……(c.s.c.t.) seg BP seg CR ...( and ∠PAQ ∠RCQ ∴ seg AP || seg CR … AP = PB)…(1) ……(c.a.c.t) (alternate angles test) ∴ seg BP || seg CR …(AP = PB)....(2) 73
9011041155 / 9011031155 PBCR is a parallelogram (from (1) and (2)) ∴ side PR || side BC …..(opposite sides of a parallelogram) seg PQ || side BC ……(P - Q - P) 1 Also, seg PQ = side PR 2 …(construction) ∴ seg PQ = 1 side BC …( 2 PR = BC) Hence, In a triangle, line segment joining mid - points of any two sides is parallel to the third side and is half of it. 74
9011041155 / 9011031155 Theorem 18 Converse of mid - point theorem If a line drawn through the mid - point of one side of a triangle is parallel to second side then it bisects the third side. In ∆ABC, A - D - B if AD = DB and line DE || side BC then, line DE bisects side AC. i.e. AE = CE Try to write the proof. 75
9011041155 / 9011031155 Sums 14. In the figure, □PQRS and □LMNR are rectangles, where M is the midpoint of PR. Prove that: i. SL = LR ii. LN = ½ SQ. Proof : In above fig. □LMNR is a rectangle. ∴ LM || RN … (Opposite sides of a rectangle) i.e. LM || RQ … (R-N-Q) … (1) 76
9011041155 / 9011031155 RQ || SP … (opposite sides of a rectangle) …(2) From (1) and (2), we have LM || RQ || SP … (3) In PSR, it is given that M is the midpoint of PR and LM || RQ … [From (1)] ∴ By the converse of midpoint theorem, point L is the midpoint of side SR. ∴ SL = LR 77 … (4) …i
9011041155 / 9011031155 Again, LR || MN and LR || PQ … (opposite sides of rectangle LMNR and rectangle PQRS respectively) ∴ MN || PQ … (property for parallel sides) ∴ In RPQ, M is the midpoint of PR. 78
9011041155 / 9011031155 ∴ By the converse of midpoint theorem, point N is the midpoint of RQ. … (5) From (4) and (5), L and N are the midpoints of SR and RQ respectively., ∴ In RSQ, by midpoint theorem, LN = ½ SQ Hence proved. 79 … ii
9011041155 / 9011031155 15. In the figure, seg PD is the median of PQR. Point PM 1 G is the midpoint of seg PD. Show that = . PR 3 Proof : Draw seg DN || seg QM. We are given that, In PDN, G is the midpoint of seg PD. Seg QM (i.e. seg GM) || seg DN … (by construction) 80
9011041155 / 9011031155 ∴ By the converse of midpoint theorem, M is the midpoint of seg PN. ∴ PM = MN … (1) In QRM, As PD is median, D is the midpoint of side QR And seg QM || seg DN … (construction) ∴ By the converse of midpoint theorem, N is the midpoint of seg MR. ∴ MN = NR 81 … (2)
9011041155 / 9011031155 From (1) and (2), PM = MN = NR … (3) Now, PR = PM + MN + NR … (4) ∴ PR = 3PM … [From (3) and (4)] i.e. 3PM = PR 82 PM 1 ∴ = . PR 3
9011041155 / 9011031155 16. Prove that a diagonal of a rhombus bisects two opposite angles. Proof : Refer above fig. ABCD is a rhombus & AC is its diagonal. In ABC and ADC, side AB side BC side AD and side DC … (sides of a rhombus) side AC side AC … (common side) ∴ ABC ADC 83 … (SSS test)
9011041155 / 9011031155 ∴ ∠BAC and ∠BCA ∠DAC … (1) ∠DCA … (2) … (c.a.c.t.) seg AC bisects ∠BAD. …[from(1)] seg AC bisects ∠BCD. …[From(2)] Therefore diagonal of a rhombus bisects two opposite angels. 84
9011041155 / 9011031155 17. In the figure, BHIR is a kite. If ∠HIR = 50º, find a. ∠HRI and b. ∠BHI. Sol. Consider abive fig. □BHIR is a kite. a. In IRH, side IR = side IH ∴ ∠HRI = ∠RHI … (given) … (angles opposite to equal sides) … (1) ∠I + ∠HRI + ∠RHI = 180º … (angles of a triangle) 85
9011041155 / 9011031155 ∴ 50º + ∠HRI + ∠HRI = 180º … [Given and from (1)] ∴ 2∠HRI = 180º - 50º ∴ 2∠HRI = 130º ∴ ∠HRI = 65º … (2) ∴ ∠RHI = 65º … [From (1) and (2)] … (3) 86
9011041155 / 9011031155 b. BHR is an equilateral triangle. … (given) ∴ ∠BHR = 60º … (angle of an equilateral triangle) … (4) ∠BHI = ∠BHR + ∠RHI … (angle addition postulate) = 60º + 65º … [From (4) and (3)] ∴ ∠BHI = 125º 87
9011041155 / 9011031155 18. In CBS, seg BC seg SC. Ray CE bisects exterior ∠DCS. Ray SE || ray BC. Prove that CBSE is a parallelogram. Proof : We are given that seg CB ∴ ∠CBS seg CS ∠CSB … (angles opposite to congruent sides) … (1) ∠DCS is the exterior angle of CBS. 88
9011041155 / 9011031155 ∴ ∠DCS = ∠CBS + ∠CSB … (Remote interior angle theorem) … (2) From (1) and (2), . ∠CBS = ∠CSB = ½∠DCS … (3) Now, it is given that Ray CE bisects ∠DCS. ∴ ∠DCE = ∠ECS = ½∠DCS … (4) 89
9011041155 / 9011031155 From (3) and (4), ∠CSB = ∠ECS ∴ seg BS || seg CE … (Alternate angles test for parallel lines) seg SE || seg BC … (given) ∴ Both the pairs of opposite sides of ∴ CBSE are parallel. CBSE is a parallelogram. 90
9011041155 / 9011031155 19. In the figure, LAXM is a parallelogram. Point I is the midpoint of diagonal LX. PQ is a line passing through I. P and Q are the points of intersection with the sides LA and MX respectively. Prove that seg PI Proof : seg IQ. In LPI and XQI, ∠LIP & ∠XIQ are vertically opposite angles ∴∠LIP ∠XIQ seg LI seg XI … (As I is the midpoint of LX) 91
9011041155 / 9011031155 ∠PLI ∠QXI … (Alternate angles) ∴ LPI XQI ∴ seg PI seg QI Hence proved. 92 … (ASA test) … (c.s.c.t.)
9011041155 / 9011031155 20. MNOP is a rhombus. Q is a point in the interior of the rhombus such that QM = QO. Prove that Q lies on diagonal NP. Proof : In MPN and OPN, seg PM seg PO and seg MN seg ON … (sides of a rhombus) seg PN sg PN … (common side) ∴By SSS test ∴ MPN OPN 93
9011041155 / 9011031155 ∴ ∠MPN ∠OPN … (c.a.c.t.) i.e. diagonal PN is the bisector of ∠MPO … (1) In MPQ and OPQ, seg MP seg OP … (sides of a rhombus) seg QM sg QO … (given) seg PQ seg PQ … (common side) ∴ By SSS test ∴ MPQ OPQ ∴∠MPQ ∠OPQ 94 … (c.a.c.t.)
9011041155 / 9011031155 i.e. seg PQ is the bisector of ∠MPO … (2) Point Q is in the interior of MNOP. … (3) The bisector of an angle is unique. ∴From (1), (2) and (3) bisectors PN and PQ of ∠MPO are one and the same. ∴ Point Q lies on the diagonal NP. Hence proved. 95
9011041155 / 9011031155 21. ABCD is a square. P and Q are the points such that seg AQ seg DP. Prove that seg AQ ⊥ seg DP. Proof : In DAP and ABQ, ∠DAP ∠ABQ … (Each is a right angle) It is given that hypotenuse DP hypotenuse AQ side DA ≅ side AB … (side of a square) ∴By hypotenuse-side theorem, 96
9011041155 / 9011031155 ∴ DAP ABQ ∴ ∠DPA ∠AQB … (c.a.c.t.) and ∠ADP … (1) ∠BAQ … (c.a.c.t.) … (2) From (1), Assume ∠DPA = ∠AQB = α … (3) From (2), Assume ∠ADP = ∠BAQ = β 97 … (4)
9011041155 / 9011031155 Then, In DAP, α+β+ DAP = 180º … (Angles of a triangle) ∴ α + β + 90º = 180º … (∠DAP = 90º) ∴ α +β = 90º … (5) Now, In APT, α + β + ∠ATP = 180º … (Angles of a triangle) ∴ 90º + ∠ATP = 180º … [From (5)] ∴ ∠ATP = 90º 98
9011041155 / 9011031155 Point T is the point of intersection of seg AQ and seg PD. ∴ seg AQ ⊥ seg DP. Hence proved!! 99
9011041155 / 9011031155 22. ABCD is a kite. AB = AD and CB = CD. Prove that i. diagonal AC ⊥ diagonal BD. ii. diagonal AC bisects diagonal BD. Proof : We are given, AB = AD ∴ Point A is equidistant from points B and D of seg BD. Also, CB = CD … (1) … (given) ∴ Point C is equidistant from points B and D of seg BD. 100 … (2)
9011041155 / 9011031155 From (1) and (2), points A and C are equidistant from points B and D of seg BD. Therefore by perpendicular bisector theorem, AC is the perpendicular bisector of BD. i.e. i. diagonal AC ⊥ diagonal BD and ii. diagonal AC bisects diagonal BD. Hence proved!! 101
9011041155 / 9011031155 23. Let points A and B be on one side of line ℓ. Draw seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C be the midpoint of seg AB. Prove that seg CD seg CE. Construction : Draw seg CM ⊥ lin l. Proof Refer above fig. : We can have, Seg AD, seg BE and seg CM are perpendiculars to line l ∴ seg AD || seg BE || seg CM. 102
9011041155 / 9011031155 Seg CA seg CB … (As C is the midpoint of seg AB) Now, Seg CA and seg CB are the intercepts made by three parallel lines AD, BE and CM. As DM & ME are intercepts made by the same parallel lines on other transversal, we have ∴ intercept DM 103 intercept ME
9011041155 / 9011031155 i.e. seg DM seg ME … (1) In CDM and CEM, seg DM seg EM … [From (1)] ∠CMD ∠CME … (Each a right angle : construction) seg CM seg CM … (common side) ∴ CDM CEM … (SAS test) ∴ seg CD 104 seg CE … (c.s.c.t.)
9011041155 / 9011031155 24. □ABCD is a parallelogram. P, Q, R and S are the points on sides AB, BC, CD and DA respectively such that seg AP Prove that Proof : seg BQ seg CR seg DS. PQRS is a parallelogram. AB = AP + PB … (A-P-B) … (1) CD = CR + RD … (C-R-D) … (2) side AB side CD … (opposite sides of a parallelogram) ∴ AB = CD … (3) 105
9011041155 / 9011031155 From (1), (2) and (3), AP + PB = CR + RD … (4) seg AP … (given) seg CR ∴ AP = CR … (5) From (4) and (5), PB = RD ∴ seg PB seg RD … (6) In PBQ and RDS, seg PB seg RD … [From (6)] Now, ∠B &∠D are opposite angles of a parallelogram ∴∠B ∠D 106
9011041155 / 9011031155 seg BQ seg DS … (given) ∴By SAS test ∴ PBQ RDS ∴ seg PQ ≅ seg RS … (c.s.c.t.) Similarly, seg PS proved. seg QR can be … (8) From (7) and (8), opposite sides of PQRS are congruent. ∴ PQRS is a parallelogram. Hence proved!! 107
9011041155 / 9011031155 Ask Your Doubts call For inquiry and registration, call 9011041155 / 9011031155. 108

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9th Maths - Quadrilateral and Its Types

  • 1. 9011041155 / 9011031155 • Live Webinars (online lectures) with recordings. • Online Query Solving • Online MCQ tests with detailed solutions • Online Notes and Solved Exercises • Career Counseling 1
  • 2. 9011041155 / 9011031155 Quadrilaterals We know that quadrilateral is a closed figure with four sides. But by definition suppose P, Q, R and S are four points in a plane such that i. 3 of these points are not collinear ii. If sides PQ, QR, RS, SP are such that any of the two such segments have a common point, it is an endpoint only. iii. If we draw a line containing any one of these segments the remaining two points lie on the same side of this line and iv. If seg PR and seg QS intersect in the points other than P, Q, R, S then the union of 4 segments PQ, QR, RS, SP is called a quadrilateral. is a symbol of ‘quadrilateral’. 2
  • 3. 9011041155 / 9011031155 (And Interior of the quadrilateral is a convex set but quadrilateral is not a convex set) Each quadrilateral has 4 sides, 4 angles, 4 vertices, 2 diagonals, 4 pairs of adjacent angles, 2 pairs of opposite sides, 4 pairs of adjacent or consecutive sides and 2 pairs of opposite angles. There are main 8 types of quadrilaterals (as shown in the figures they are i. Parallelogram ii. Rectangle iii. Rhombus iv. Square v. Trapezium (non parallel sides are not congruent) vi. Isosceles trapezium vii. Kite viii. Kite (with all side, congruent) 3
  • 4. 9011041155 / 9011031155 Table S. No. Quadrilateral Type 1. Parallelogram 2. Rectangle 3. Rhombus 4. Square 4
  • 5. 9011041155 / 9011031155 5. Trapezium 6. Isosceles Trapezium 7. Kite I 8. Kite II 5
  • 6. 9011041155 / 9011031155 Property of a quadrilateral Theorem 1 [Angle sum property of a quadrilateral] The sum of the measures of all angles of a quadrilateral is 360º Given : To prove : Construction : ABCD is a given quadrilateral ∠A + ∠B + ∠C + ∠D = 360º Join the points A and C 6
  • 7. 9011041155 / 9011031155 Proof : In ABC, ∠BAC + ∠ ABC + ∠BCA = 180º .....sum of angles of ABC .... (1) and, In ADC, ∠DAC + ∠ADC + ∠DCA = 180º .....sum of angles of ADC .... (2) 7
  • 8. 9011041155 / 9011031155 Adding (1) and (2) we get ∠BAC + ∠ABC + ∠BCA + ∠DAC + ∠ADC + ∠DCA = 180º + 180º ∴ ∠BAC + ∠DAC + ∠ABC + ∠BCA + ∠DCA + ∠ADC = 360º ......(3) But ∠BAC + ∠DAC = ∠BAD ....Angle addition property....(4) and ∠BCA + ∠DCA = ∠BCD ....Angle addition property....(5) ∴ from (3), (4) and (5) we get 8
  • 9. 9011041155 / 9011031155 ∠BAD + ∠ABC + ∠BCD + ∠ADC = 360º i.e. ∠A + ∠B + ∠C + ∠D = 360º Hence, sum of the measures of all angles of a quadrilateral is 360º 9
  • 10. 9011041155 / 9011031155 Theorem 2 The opposite sides of a parallelogram are congruent. Given ABCD is a parallelogram in which : side AB || side CD and side BC || side DA : Construction : Proof : side AB side CD and side BC To prove side DA Draw diagonal AC. ABCD is a parallelogram in which side AB || side CD and seg AC is a transversals. 10
  • 11. 9011041155 / 9011031155 ∴∠BAC ∠DCA ...(alternate angle)....(1) Similarly, side BC || side DA and seg AC is a transversal. ∴ ∠BCA ∠DAC ......(alternate angle) ...(2) In ABC and ∠BAC seg AC ∠BCA ∴ CDA ∠DCA ......(from 1) seg CA .......(common side) ∠ DAC ABC CDA 11 ........(from 2) .......(ASA test)
  • 12. 9011041155 / 9011031155 ∴ side AB side CD and side BC ......(c.s.c.t) side DA ......(c.s.c.t.) Hence, the opposite sides of a parallelogram are congruent. 12
  • 13. 9011041155 / 9011031155 Theorem 3 If opposite sides of a quadrilateral are congruent then the quadrilateral is parallelogram. Given : In ABCD side AB side DC and side AD To prove : Construction : side BC ABCD is a parallelogram Draw a diagonal BD 13
  • 14. 9011041155 / 9011031155 Proof : In ABD and CDB side AB side DC side AD side BC side BD side DB ......(given) ....(common side) ∴ ABD CDB ∴ ∠ABD ∠CDB .....(SSS test) ......(c.a.c.t.) ∴ side AB || side CD .......(1) ...(alternate angles test) 14
  • 15. 9011041155 / 9011031155 Similarly, we can prove that side AD || side BC ......(2) ....(alternate angels test) ∴ from (1) and (2) we have ABCD is a parallelogram Hence, if opposite sides of a quadrilateral are congruent then it is a parallelogram. 15
  • 16. 9011041155 / 9011031155 Theorem 4 The opposite angles of a parallelogram are congruent Given PQRS is a parallelogram : To prove : ∠SPQ ∠QRS and ∠PSR Construction : Proof : ∠RQP Draw a diagonal SQ PQRS is a parallelogram ∴ side PS || side QR and seg SQ is a transversal ∴ ∠PSQ ∠RQS ..(Alternate angles) ...(1) 16
  • 17. 9011041155 / 9011031155 Also, side PQ || side SR and seg SQ is a transversal. ∴ ∠PQS ∠RSQ ...(Alternate angles)..(2) In PQS and ∠PSQ RSQ ∠RQS ......from (1) side SQ side QS ....(common side) ∠PQS ∴ ∠RSQ PQS ......from (2) RSQ ......(ASA test) ∴ ∠SPQ 17 ∠QRS ......(c.a.c.t.)
  • 18. 9011041155 / 9011031155 Similarly, we can prove by drawing diagonal PR. ∠PSR ∠ RQP Hence, the opposite angles of a parallelogram are congruent. 18
  • 19. 9011041155 / 9011031155 Theorem 5 A quadrilateral is a parallelogram if its opposite angles are congruent. Given : PQRS is a parallelogram in which ∠SPQ ∠PQR To prove : Proof : ∠QRS and ∠RSP PQRS is a parallelogram Let ∠SPQ = ∠QRS = xº And ∠PQR = ∠RSP = yº ......Opposite angle of a quadrilateral. 19
  • 20. 9011041155 / 9011031155 ∠SPQ + ∠PQR + ∠QRS + ∠RSP= 360º. …(Angle sum property of a quadrilateral) ∴ x + y + x + y = 360º ∴ 2x + 2y = 360º ∴ x + y = 180º .....(dividing by 2) ∴ ∠SPQ + ∠RSP = 180º ∴ side PQ || side SR ... (interior angles test) ....(1) 20
  • 21. 9011041155 / 9011031155 Similarly, we can prove that side PS || side QR .......(2) ∴ PQRS is a parallelogram ...... from (1) and (2) Hence, if opposite angles of a quadrilateral are congruent, then it is a parallelogram. 21
  • 22. 9011041155 / 9011031155 Theorem 6 The diagonals of a parallelogram bisect each other. Given : ABCD is a parallelogram in which the diagonals AC and BD intersect in M. To prove : seg AM seg CM and seg BM Proof : since seg DM ABCD is a parallelogram side AB || side CD and AC is a transversal. ∴ ∠BAC ∠DCA ...(Alternate angles) i.e. ∠BAM ∠DCM ...(A – M – C)..(1) 22
  • 23. 9011041155 / 9011031155 Also, side AB || side DC and seg DB is a transversal. ∴ ∠ABD ∠CDB ...(Alternate angles) i.e. ∠ABM ∠CDM ...(B – M – D) ..(2) Now, In ABM and ∠BAM ∠DCM ......(from 1) side AB ∠ABM ∴ CDM side DC .....(opposite sides) ∠CDM ......(from 2) ABM CDM ......(ASA test) 23
  • 24. 9011041155 / 9011031155 ∴ seg AM seg CM and seg BM .....(c.s.c.t.) seg DM Hence, diagonals of a parallelogram bisect each other. 24
  • 25. 9011041155 / 9011031155 Theorem 7 If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. Given : PQRS is a quadrilateral in which diagonals PR and QS intersect in M. seg PM seg QM To prove : Proof : seg RM and seg SM PQRS is a parallelogram In PMQ and seg PM RMS seg RM 25 ......(given)
  • 26. 9011041155 / 9011031155 ∠PMQ ∠RMS .....(verticallyopposite angles) seg QM ∴ seg SM .....(given) PMQ RMS ∴ ∠PQM ∠RSM i.e. ∠PQS ......(SAS test) ......(c.a.c.t) ∠RSQ ......(S – M – Q) ∴ side PQ || side SR ....(alternate angles test) ...(1) Similarly, we can prove that side PS || side QR 26 .....(2)
  • 27. 9011041155 / 9011031155 PQRS is a parallelogram .........from (1) and (2) Hence, if the diagonals of a quadrilateral bisect each other then it is a parallelogram. 27
  • 28. 9011041155 / 9011031155 Theorem 8 A quadrilateral is a parallelogram if a pair of opposite sides is parallel and congruent. Given LMNK is a given quadrilateral in : which side LM || side NK and side LM To prove : Construction : side NK LMNK is a parallelogram Draw diagonal MK 28
  • 29. 9011041155 / 9011031155 Proof : since LMNK is a quadrilateral in which side LM || side NK and seg MK is a transversal. ∴∠LMK ∠NKM ..(Alternate angles) ...(1) Now, In KLM and MNK seg LM seg NK .....(given) ∠LMK ∠NKM seg KM .......(from 1) seg MK .......(common side) ∴ KLM MNK ......(SAS test) 29
  • 30. 9011041155 / 9011031155 ∴ ∠LKM ∠NMK .......(c.a.c.t) ∴ side LK || side MN .....(alternate angles test) ....(2) and side LM || side NK ...(given) .....(3) ∴ from (2) and (3) we have LKMN is a parallelogram Hence, if a pair of opposite sides of a quadrilateral is parallel and congruent then the quadrilateral is a parallelogram. 30
  • 31. 9011041155 / 9011031155 Sums 1. The angles of a quadrilateral are in the ratio 2 : 3 : 5 : 8. Find all the angles of the quadrilateral. Sol. The angles of the quadrilateral are in ratio 2 : 3 : 5 : 8. Let the measures of the four angles are 2xº, 3xº, 5xº and 8xº. According to the angle sum property of a quadrilateral, we have 2xº + 3xº + 5xº + 8xº = 360º ∴ 18xº = 360º 3600 ∴x = 18 0 ∴ x 0 = 200 2xº = 2 × 20º = 40º, 3xº = 3 × 20º = 60º, 5xº = 5 × 20º = 100º and 8xº = 8 × 20º = 160º ∴ The measures of the angles of the quadrilateral are 40º, 60º, 100º and 160º. 31
  • 32. 9011041155 / 9011031155 2. In the figure, PQRS is a quadrilateral. The bisectors of ∠P and ∠Q meet at the point A. ∠S = 50º, ∠R = 110º. Find the measure of∠PAQ. Sol. Refer above fig. According to the angle sum property of a quadrilateral, ∠SPQ + ∠PQR + ∠R + ∠S = 360º ∴ ∠SPQ + ∠PQR + 110º + 50º = 360º … (substituting the given values) 32
  • 33. 9011041155 / 9011031155 ∴ ∠SPQ + ∠PQR + 160º = 360º ∴ ∠SPQ + ∠PQR = 360º - 160º ∴ ∠SPQ + ∠PQR = 200º … (1) Now, it is given that Ray PA and ray QA are the bisectors of ∠SPQ and ∠RQP respectively. ∴ ∠APQ = ½ ∠SPQ and ∠AQP = ½ ∠RQP ∴ ∠APQ + ∠AQP = ½ (∠SPQ + ∠RQP) … (2) From (2) and (1), ∠APQ + ∠AQP = ½ (200º) ∴ ∠APQ + ∠AQP = 100º 33 … (3)
  • 34. 9011041155 / 9011031155 Also, In APQ, ∠PAQ + ∠APQ + ∠AQP = 180º … (Angles of a triangle) ∴ ∠PAQ + 100º = 180º … [From (3)] ∴ ∠PAQ = 180º - 100º ∴ ∠PAQ = 80º ∴The measure of ∠PAQ is 80º. 34
  • 35. 9011041155 / 9011031155 3. The sides BA and DC of quadrilateral ABCD are extended as shown in the figure. Prove that m + n = p + q. Proof : Consider ∠DAB b = a and ∠BCD = c. According to the angle sum property of a quadrilateral, In quadrilateral ABCD, a + q + c + p = 360º 35 … (1)
  • 36. 9011041155 / 9011031155 From (2), m + c + n + a = 360º … (3) From (1) and (3), a+q+c+p=m+c+n+a ∴q+p=m+n i.e. m + n = p + q. 36
  • 37. 9011041155 / 9011031155 4. In a parallelogram ABCD, ∠A = xº, ∠B = (3x + 20)º. Find x, ∠C and ∠D. Sol. We know that the adjacent angles of a parallelogram are supplementary. ∴ ∠A + ∠B = 180º Put the given values ∴ xº + 3xº + 20º = 180º ∴ 4xº = 180º - 20º ∴ 4xº = 160º 37
  • 38. 9011041155 / 9011031155 ∴ xº = 40º 3xº + 20º = 3 × 40º + 20º = 120º + 20º = 140º ∴ ∠A = 40º and ∠B = 140º Also, the opposite angles of a parallelogram are congruent. ∴ ∠C = ∠ A = 40º and ∠ D = ∠ B = 140º ∴ x = 40. ∠ C = 40º, ∠ D = 140º. 38
  • 39. 9011041155 / 9011031155 5. The perimeter of a parallelogram is 150 cm. One of its sides is greater than the other by 25 cm. Find the lengths of all the sides of the parallelogram. Sol: Consider that the smaller of the two adjacent sides be x cm. Then, it is given that the other side is (x + 25) cm Now, the perimeter of a parallelogram is equal to the sum of the lengths of four sides. ∴ x + (x + 25) + x + (x + 25) = 150 ∴ 4x + 50 = 150 ∴ x = 25 and x + 25 = 25 + 25 = 50 ∴ The lengths of the sides of the parallelogram ∴ 4x = 150 – 50 ∴ 4x = 100 are 25 cm, 50 cm, 25 cm and 50 cm. 39
  • 40. 9011041155 / 9011031155 6. In the figure, WXYZ is a parallelogram. From the information given in the figure, find the values of x and y. Sol. As XY & ZW are opposite sides of a parallelogram side XY || side ZW seg XZ is a transversal. ∴ ∠ ZXY = ∠ XZW ∴ 4y = 28º … (alternate angles) … (given) ∴ y = 7º 40
  • 41. 9011041155 / 9011031155 Similarly, side XW || side YZ and seg XZ is the transversal. ∴ ∠ ZXW = ∠ XZY. ∴ 10x = 60º … (given) ∴ x = 6º. ∴The value of x is 6º and that of y is 7º. 41
  • 42. 9011041155 / 9011031155 7. Prove that two opposite vertices of a parallelogram are equidistant from the diagonal not containing these vertices. Given : 1. ABCD is a parallelogram. 2. Seg AM ⊥ diagonal BD and Seg CN ⊥ diagonal BD. We have to prove seg AM ≅ seg CN. First, draw diagonal AC intersecting diagonal BD in the point P. 42
  • 43. 9011041155 / 9011031155 Proof : P is the point of intersection of the diagonals of parallelogram ABCD. ∴ seg AP ≅ seg CP … (1) In APM and CPN, seg AP ≅ seg CP … [From (1)] ∠ APM ≅ ∠ CPN … (vertically opposite angles) ∠ AMP ≅ ∠CNP … (each a right angle) 43
  • 44. 9011041155 / 9011031155 ∴ APM ≅ CPN … (SAA test) ∴ seg AM ≅ seg CN … (c.s.c.t.) Therefore, the opposite vertices of a parallelogram are at equal distance from the diagonal not containing these vertices. 44
  • 45. 9011041155 / 9011031155 8. The ratio of two sides of a parallelogram is 3 : 4. If its perimeter is 112 cm, find the lengths of the sides of the parallelogram. Sol: It is given that the ratio of the lengths of two sides of the parallelogram is 3:4. Let the lengths of those two sides be 3x cm and 4x cm respectively. We know that the opposite sides of a parallelogram are congruent. ∴ The sides of the parallelogram are 3x cm, 4x cm, 3x cm and 4x cm. Now, the perimeter of a parallelogram = the sum of the lengths of four sides ∴ 112 = 3x + 4x + 3x + 4x … (given perimeter 112 cm) 45
  • 46. 9011041155 / 9011031155 ∴ 14x = 112 ∴ x=8 3x = 3 × 8 = 24 and 4x = 4 × 8 = 32 ∴ The lengths of the sides of the given parallelogram are 24 cm, 32 cm, 24 cm and 32 cm. 46
  • 47. 9011041155 / 9011031155 Definitions Parallelogram A quadrilateral is a parallelogram if its opposite sides are parallel. Rectangle A parallelogram (or a quadrilateral) in which each angle is a right angle, is called a rectangle. Rhombus A quadrilateral having all sides congruent is called a rhombus. Square A quadrilateral is called a square if all its sides and angles are congruent. 47
  • 48. 9011041155 / 9011031155 Trapezium If only one pair of opposite sides is parallel, the quadrilateral is said to be a trapezium. Isosceles trapezium A trapezium in which nonparallel sides are congruent, it is called as isosceles trapezium. Kite If in ABCD, AB AD and CB CD and diagonal AC is perpendicular bisector of diagonal BD, then ABCD is called as a kite. 48
  • 49. 9011041155 / 9011031155 Theorem 9 Diagonals of a rectangle are congruent. Theorem 10 If diagonals of a parallelogram are congruent then it is a rectangle. 49
  • 50. 9011041155 / 9011031155 Theorem 11 Diagonals of a rhombus are perpendicular bisectors of each other. Theorem 12 If the diagonals of a quadrilateral bisect each other at right angle then the quadrilateral is a rhombus. 50
  • 51. 9011041155 / 9011031155 Theorem 13 Diagonals of a square are congruent and perpendicular bisectors of each other. Theorem 14 If diagonals of quadrilateral are congruent and perpendicular bisectors of each other then it is square. 51
  • 52. 9011041155 / 9011031155 Sums 1. Adjacent sides of a rectangle are of lengths 7 cm and 24 cm. Find the lengths of its diagonals. Sol. Consider ABCD as a rectangle in which AB = 24 cm and BC = 7 cm. In right-angled ABC, By Pythagoras’ theorem, AC² = AB² + BC² = (24)² + (7)² = 576 + 49 = 625 ∴ AC = 25 52
  • 53. 9011041155 / 9011031155 But, the diagonals of a rectangle are congruent. ∴ BD = AC = 25 ∴ The length of diagonal AC = 25 cm and that of BD = 25 cm. 53
  • 54. 9011041155 / 9011031155 2. ABCD is a trapezium in which AB || DC. M and N are the midpoints of side AD and side BC respectively. If AB = 12 cm and MN = 14 cm, find CD. Sol. The length of the segment joining the midpoints of non - parallel sides of a trapezium is half the sum of the lengths of its parallel sides. ∴ MN = ½ (AB + CD) ∴ 14 = ½ (12 + CD) … (substituting the given values) 54
  • 55. 9011041155 / 9011031155 ∴ 28 = 12 + CD … (Multiplying both the sides by 2) ∴ CD = 28 – 12 ∴ CD = 16. 55
  • 56. 9011041155 / 9011031155 3. Prove that the base angles of an isosceles trapezium are congruent. Given : □ ABCD is an isosceles trapezium in which side AB || side CD and side AD To prove : Construction : ∠ADC side BC. ∠BCD. Draw seg BE || side AD and D-E-C. 56
  • 57. 9011041155 / 9011031155 Proof : IN □ABED, side AB || side DE. … (given) seg BE || side AD … (construction ) ∴ □ABED is a parallelogram. The opposite sides and the opposite angles of a parallelogram are congruent. ∴ side AD and ∠D 57 side BE … (1) ∠ABE … (2)
  • 58. 9011041155 / 9011031155 side AD ≅ side BC … (given) … (3) From (1) and (3), side BE ≅ side BC ∴ ∠ BEC ≅ ∠C …(Isosceles triangle theorem) … (4) side AB || side DC … (given) side BE is the transversal. ∴ ∠ABE ≅ ∠BEC … (alternate angles) … (5) From (2), (4) and (5), ∠D ≅ ∠C. i.e. ∠ADC ≅ ∠BCD. [Similarly, we can prove that ∠DAB 58 ∠CBA.]
  • 59. 9011041155 / 9011031155 4. The lengths of the diagonals PR and QS of a rhombus are 20 cm and 48 cm respectively. Find the length of each side of the rhombus. Sol. We know that the diagonals of a rhombus bisect each other at right angles. ∴ MQ = ½ QS = ½ × 48 cm ∴ MQ = 24 cm MP = ½ PR = ½ × 20 cm ∴ MP = 10 cm. ∠ PMQ = 90º 59
  • 60. 9011041155 / 9011031155 In right-angled PMQ, By Pythagoras’ theorem PQ² = PM² + QM² = (10)² + (24)² = 100 + 576 = 676 ∴ PQ = 26. But all the sides of a rhombus are congruent. ∴ The length of each side of the given rhombus is 26 cm. 60
  • 61. 9011041155 / 9011031155 5. ABCD is a kite in which diagonal AC and diagonal BD intersect at point O. ∠ OBC = 20º and ∠OCD = 40º. Find : i. ∠ABC ii. ∠ADC iii. ∠BAD. Sol. Refer above fig, The diagonal joining the unequal sides of a kite bisects the other diagonal at right angles. ∴ Each angle at O is a right angle. Also, side BA side BC. 61
  • 62. 9011041155 / 9011031155 Side DA side DC and seg AO seg CO … (perpendicular bisector theorem) … (1) In BOC, ∠BOC = 90º, ∠OBC = 20º … (given) ∴ ∠OCB = 180º - (90º + 20º) = 180º - 110º ∴ ∠OCB = 70º … (2) 62
  • 63. 9011041155 / 9011031155 In BAC, BA = BC … [From (1)] ∴ ∠BCA = ∠BAC … (angles opposite to equal sides) i.e. ∠OCB = ∠OAB ∴ ∠OAB = 70º … [From (2)] In DAC, DA = DC ∴ ∠DCA = ∠DAC … [From (1)] … (angles opposite to equal sides) i.e. ∠DCO = ∠DAO 63 … (3)
  • 64. 9011041155 / 9011031155 ∴ ∠DAO = 40º … [Given : ∠DCO = 40º] … (4) From (3) and (4), ∠OAB + ∠DAO = 70º + 40º = 110º … (5) ∠BAO + ∠DAO = ∠BAD … (Angle addition postulate) … (6) From (5) and (6), ∠BAD = 110º. In DAC, ∠ADC + ∠DAC + ∠DCA = 180º … (angles of a triangle) ∴ ∠ ADC + 40º + 40º = 180º … [From (4) and given] 64
  • 65. 9011041155 / 9011031155 ∴ ∠ADC = 180º - 80º ∴ ∠ADC = 100º … (7) In BAC, ∠ABC + ∠BAC + ∠BCA = 180º … (angles of a triangle) ∴ ∠ABC + 70º + 70º = 180º … [From (2) and 3)] ∴ ∠ABC = 180º - 140º ∴ ∠ABC = 40º Ans. : i. ∠ABC = 40º ii. ∠ADC = 100º iii. ∠BAD = 110º. 65
  • 66. 9011041155 / 9011031155 Theorem 15 Diagonals of an isosceles trapezium are congruent. Given ABCD is an isosceles trapezium in : which seg AC and seg BD are diagonals side AD side BC and side AB || side DC To Prove : Construction : seg AC seg BD Draw seg AM ⊥ seg DC, seg BN ⊥ seg DC Proof : In ∆AMD and ∆BNC side AD side BC 66 ….(given)
  • 67. 9011041155 / 9011031155 ∠AMD seg AM ∠BNC …..(90º each) seg BN …..(perpendicular distance between two parallel lines) ∴ ∆AMD ∆BNC ……(hypotenuse side test) ∴ ∠ADM ∠BCN …..(c. a. c. t.) i.e. ∠ADC ∠BCD .(D - M - C)…(1) Now, In ∆ADC and ∆BCD side AD ∠ADC side BC ∠BCD 67 ….(given) …..(from (1))
  • 68. 9011041155 / 9011031155 side DC ∴ ∆ADC side CD …..(common side) ∆BCD ∴ seg AC …..(SAS test) seg BD ……(c.s.c.t.) Hence, diagonals of an isosceles trapezium are congruent. 68
  • 69. 9011041155 / 9011031155 Theorem 16 (Intercepts made, by three parallel lines) If three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. Given : line || line m || line n and lines t1 and t2 are transversals. seg AB Prove : Construction : seg BC seg DE seg EF Draw a line GI parallel to the line t1 through E. 69
  • 70. 9011041155 / 9011031155 Proof : seg AG || seg BE ……(given) seg AB || seg GE ……. (construction) ∴ ABEG is a parallelogram. ∴ seg AB seg GE…(opposite sides of a parallelogram …(1) Similarly, ∴ seg BC BCIE is a parallelogram. seg EI …..(opposite sides of a parallelogram….(2) But seg AB seg BC ….. (given)...(3) Hence seg GE seg EI ….(4) from (1), (2) and (3) 70
  • 71. 9011041155 / 9011031155 Now, In ∆GED and ∆IEF ∠DGE ∠FIE …(alternate angles) seg GE ∠GED seg EI …..(from (4)) ∠IEF ……(vertically opposite angles) ∴ ∆GED ∴ seg DE ∆IEF ……(ASA test) seg EF ……(c.s.c.t.) Hence, if three parallel lines make congruent intercepts on a transversal then they make congruent intercepts on any other transversal. 71
  • 72. 9011041155 / 9011031155 Theorem 17 Mid - Point theorem The line segment joining the midpoints of any two sides of a triangle is parallel to the third side and is half of it. Given : In ∆ABC, P and Q are midpoints of sides AB and AC respectively. To prove : i. seg PQ || side BC ii. seg PQ = 1 side BC 2 Construction : Take the point R on ray PQ such that seg PQ seg QR. Join C and R 72
  • 73. 9011041155 / 9011031155 Proof : In ∆AQP and ∆CQR seg AQ ∠AQP seg CQ ……(given) ∠CQR ….(vertically opposite angles) seg PQ seg RQ ……(construction) ∴ ∆AQP ∆CQR …… (SAS test) seg AP seg CR ……(c.s.c.t.) seg BP seg CR ...( and ∠PAQ ∠RCQ ∴ seg AP || seg CR … AP = PB)…(1) ……(c.a.c.t) (alternate angles test) ∴ seg BP || seg CR …(AP = PB)....(2) 73
  • 74. 9011041155 / 9011031155 PBCR is a parallelogram (from (1) and (2)) ∴ side PR || side BC …..(opposite sides of a parallelogram) seg PQ || side BC ……(P - Q - P) 1 Also, seg PQ = side PR 2 …(construction) ∴ seg PQ = 1 side BC …( 2 PR = BC) Hence, In a triangle, line segment joining mid - points of any two sides is parallel to the third side and is half of it. 74
  • 75. 9011041155 / 9011031155 Theorem 18 Converse of mid - point theorem If a line drawn through the mid - point of one side of a triangle is parallel to second side then it bisects the third side. In ∆ABC, A - D - B if AD = DB and line DE || side BC then, line DE bisects side AC. i.e. AE = CE Try to write the proof. 75
  • 76. 9011041155 / 9011031155 Sums 14. In the figure, □PQRS and □LMNR are rectangles, where M is the midpoint of PR. Prove that: i. SL = LR ii. LN = ½ SQ. Proof : In above fig. □LMNR is a rectangle. ∴ LM || RN … (Opposite sides of a rectangle) i.e. LM || RQ … (R-N-Q) … (1) 76
  • 77. 9011041155 / 9011031155 RQ || SP … (opposite sides of a rectangle) …(2) From (1) and (2), we have LM || RQ || SP … (3) In PSR, it is given that M is the midpoint of PR and LM || RQ … [From (1)] ∴ By the converse of midpoint theorem, point L is the midpoint of side SR. ∴ SL = LR 77 … (4) …i
  • 78. 9011041155 / 9011031155 Again, LR || MN and LR || PQ … (opposite sides of rectangle LMNR and rectangle PQRS respectively) ∴ MN || PQ … (property for parallel sides) ∴ In RPQ, M is the midpoint of PR. 78
  • 79. 9011041155 / 9011031155 ∴ By the converse of midpoint theorem, point N is the midpoint of RQ. … (5) From (4) and (5), L and N are the midpoints of SR and RQ respectively., ∴ In RSQ, by midpoint theorem, LN = ½ SQ Hence proved. 79 … ii
  • 80. 9011041155 / 9011031155 15. In the figure, seg PD is the median of PQR. Point PM 1 G is the midpoint of seg PD. Show that = . PR 3 Proof : Draw seg DN || seg QM. We are given that, In PDN, G is the midpoint of seg PD. Seg QM (i.e. seg GM) || seg DN … (by construction) 80
  • 81. 9011041155 / 9011031155 ∴ By the converse of midpoint theorem, M is the midpoint of seg PN. ∴ PM = MN … (1) In QRM, As PD is median, D is the midpoint of side QR And seg QM || seg DN … (construction) ∴ By the converse of midpoint theorem, N is the midpoint of seg MR. ∴ MN = NR 81 … (2)
  • 82. 9011041155 / 9011031155 From (1) and (2), PM = MN = NR … (3) Now, PR = PM + MN + NR … (4) ∴ PR = 3PM … [From (3) and (4)] i.e. 3PM = PR 82 PM 1 ∴ = . PR 3
  • 83. 9011041155 / 9011031155 16. Prove that a diagonal of a rhombus bisects two opposite angles. Proof : Refer above fig. ABCD is a rhombus & AC is its diagonal. In ABC and ADC, side AB side BC side AD and side DC … (sides of a rhombus) side AC side AC … (common side) ∴ ABC ADC 83 … (SSS test)
  • 84. 9011041155 / 9011031155 ∴ ∠BAC and ∠BCA ∠DAC … (1) ∠DCA … (2) … (c.a.c.t.) seg AC bisects ∠BAD. …[from(1)] seg AC bisects ∠BCD. …[From(2)] Therefore diagonal of a rhombus bisects two opposite angels. 84
  • 85. 9011041155 / 9011031155 17. In the figure, BHIR is a kite. If ∠HIR = 50º, find a. ∠HRI and b. ∠BHI. Sol. Consider abive fig. □BHIR is a kite. a. In IRH, side IR = side IH ∴ ∠HRI = ∠RHI … (given) … (angles opposite to equal sides) … (1) ∠I + ∠HRI + ∠RHI = 180º … (angles of a triangle) 85
  • 86. 9011041155 / 9011031155 ∴ 50º + ∠HRI + ∠HRI = 180º … [Given and from (1)] ∴ 2∠HRI = 180º - 50º ∴ 2∠HRI = 130º ∴ ∠HRI = 65º … (2) ∴ ∠RHI = 65º … [From (1) and (2)] … (3) 86
  • 87. 9011041155 / 9011031155 b. BHR is an equilateral triangle. … (given) ∴ ∠BHR = 60º … (angle of an equilateral triangle) … (4) ∠BHI = ∠BHR + ∠RHI … (angle addition postulate) = 60º + 65º … [From (4) and (3)] ∴ ∠BHI = 125º 87
  • 88. 9011041155 / 9011031155 18. In CBS, seg BC seg SC. Ray CE bisects exterior ∠DCS. Ray SE || ray BC. Prove that CBSE is a parallelogram. Proof : We are given that seg CB ∴ ∠CBS seg CS ∠CSB … (angles opposite to congruent sides) … (1) ∠DCS is the exterior angle of CBS. 88
  • 89. 9011041155 / 9011031155 ∴ ∠DCS = ∠CBS + ∠CSB … (Remote interior angle theorem) … (2) From (1) and (2), . ∠CBS = ∠CSB = ½∠DCS … (3) Now, it is given that Ray CE bisects ∠DCS. ∴ ∠DCE = ∠ECS = ½∠DCS … (4) 89
  • 90. 9011041155 / 9011031155 From (3) and (4), ∠CSB = ∠ECS ∴ seg BS || seg CE … (Alternate angles test for parallel lines) seg SE || seg BC … (given) ∴ Both the pairs of opposite sides of ∴ CBSE are parallel. CBSE is a parallelogram. 90
  • 91. 9011041155 / 9011031155 19. In the figure, LAXM is a parallelogram. Point I is the midpoint of diagonal LX. PQ is a line passing through I. P and Q are the points of intersection with the sides LA and MX respectively. Prove that seg PI Proof : seg IQ. In LPI and XQI, ∠LIP & ∠XIQ are vertically opposite angles ∴∠LIP ∠XIQ seg LI seg XI … (As I is the midpoint of LX) 91
  • 92. 9011041155 / 9011031155 ∠PLI ∠QXI … (Alternate angles) ∴ LPI XQI ∴ seg PI seg QI Hence proved. 92 … (ASA test) … (c.s.c.t.)
  • 93. 9011041155 / 9011031155 20. MNOP is a rhombus. Q is a point in the interior of the rhombus such that QM = QO. Prove that Q lies on diagonal NP. Proof : In MPN and OPN, seg PM seg PO and seg MN seg ON … (sides of a rhombus) seg PN sg PN … (common side) ∴By SSS test ∴ MPN OPN 93
  • 94. 9011041155 / 9011031155 ∴ ∠MPN ∠OPN … (c.a.c.t.) i.e. diagonal PN is the bisector of ∠MPO … (1) In MPQ and OPQ, seg MP seg OP … (sides of a rhombus) seg QM sg QO … (given) seg PQ seg PQ … (common side) ∴ By SSS test ∴ MPQ OPQ ∴∠MPQ ∠OPQ 94 … (c.a.c.t.)
  • 95. 9011041155 / 9011031155 i.e. seg PQ is the bisector of ∠MPO … (2) Point Q is in the interior of MNOP. … (3) The bisector of an angle is unique. ∴From (1), (2) and (3) bisectors PN and PQ of ∠MPO are one and the same. ∴ Point Q lies on the diagonal NP. Hence proved. 95
  • 96. 9011041155 / 9011031155 21. ABCD is a square. P and Q are the points such that seg AQ seg DP. Prove that seg AQ ⊥ seg DP. Proof : In DAP and ABQ, ∠DAP ∠ABQ … (Each is a right angle) It is given that hypotenuse DP hypotenuse AQ side DA ≅ side AB … (side of a square) ∴By hypotenuse-side theorem, 96
  • 97. 9011041155 / 9011031155 ∴ DAP ABQ ∴ ∠DPA ∠AQB … (c.a.c.t.) and ∠ADP … (1) ∠BAQ … (c.a.c.t.) … (2) From (1), Assume ∠DPA = ∠AQB = α … (3) From (2), Assume ∠ADP = ∠BAQ = β 97 … (4)
  • 98. 9011041155 / 9011031155 Then, In DAP, α+β+ DAP = 180º … (Angles of a triangle) ∴ α + β + 90º = 180º … (∠DAP = 90º) ∴ α +β = 90º … (5) Now, In APT, α + β + ∠ATP = 180º … (Angles of a triangle) ∴ 90º + ∠ATP = 180º … [From (5)] ∴ ∠ATP = 90º 98
  • 99. 9011041155 / 9011031155 Point T is the point of intersection of seg AQ and seg PD. ∴ seg AQ ⊥ seg DP. Hence proved!! 99
  • 100. 9011041155 / 9011031155 22. ABCD is a kite. AB = AD and CB = CD. Prove that i. diagonal AC ⊥ diagonal BD. ii. diagonal AC bisects diagonal BD. Proof : We are given, AB = AD ∴ Point A is equidistant from points B and D of seg BD. Also, CB = CD … (1) … (given) ∴ Point C is equidistant from points B and D of seg BD. 100 … (2)
  • 101. 9011041155 / 9011031155 From (1) and (2), points A and C are equidistant from points B and D of seg BD. Therefore by perpendicular bisector theorem, AC is the perpendicular bisector of BD. i.e. i. diagonal AC ⊥ diagonal BD and ii. diagonal AC bisects diagonal BD. Hence proved!! 101
  • 102. 9011041155 / 9011031155 23. Let points A and B be on one side of line ℓ. Draw seg AD ⊥ line ℓ and seg BE ⊥ line ℓ. Let point C be the midpoint of seg AB. Prove that seg CD seg CE. Construction : Draw seg CM ⊥ lin l. Proof Refer above fig. : We can have, Seg AD, seg BE and seg CM are perpendiculars to line l ∴ seg AD || seg BE || seg CM. 102
  • 103. 9011041155 / 9011031155 Seg CA seg CB … (As C is the midpoint of seg AB) Now, Seg CA and seg CB are the intercepts made by three parallel lines AD, BE and CM. As DM & ME are intercepts made by the same parallel lines on other transversal, we have ∴ intercept DM 103 intercept ME
  • 104. 9011041155 / 9011031155 i.e. seg DM seg ME … (1) In CDM and CEM, seg DM seg EM … [From (1)] ∠CMD ∠CME … (Each a right angle : construction) seg CM seg CM … (common side) ∴ CDM CEM … (SAS test) ∴ seg CD 104 seg CE … (c.s.c.t.)
  • 105. 9011041155 / 9011031155 24. □ABCD is a parallelogram. P, Q, R and S are the points on sides AB, BC, CD and DA respectively such that seg AP Prove that Proof : seg BQ seg CR seg DS. PQRS is a parallelogram. AB = AP + PB … (A-P-B) … (1) CD = CR + RD … (C-R-D) … (2) side AB side CD … (opposite sides of a parallelogram) ∴ AB = CD … (3) 105
  • 106. 9011041155 / 9011031155 From (1), (2) and (3), AP + PB = CR + RD … (4) seg AP … (given) seg CR ∴ AP = CR … (5) From (4) and (5), PB = RD ∴ seg PB seg RD … (6) In PBQ and RDS, seg PB seg RD … [From (6)] Now, ∠B &∠D are opposite angles of a parallelogram ∴∠B ∠D 106
  • 107. 9011041155 / 9011031155 seg BQ seg DS … (given) ∴By SAS test ∴ PBQ RDS ∴ seg PQ ≅ seg RS … (c.s.c.t.) Similarly, seg PS proved. seg QR can be … (8) From (7) and (8), opposite sides of PQRS are congruent. ∴ PQRS is a parallelogram. Hence proved!! 107
  • 108. 9011041155 / 9011031155 Ask Your Doubts call For inquiry and registration, call 9011041155 / 9011031155. 108