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ACTIVE MATHS 4 BOOK 2
Given: A Δ 𝐴𝐵𝐶 and a line 𝑋𝑌 parallel to 𝐵𝐶 which cuts 𝐴𝐵 in the ratio 𝑠 ∶ 𝑡.
To prove: |AY| : |YC| = s : t.
Construction: Divide [AX] into s equal parts
and [XB] into t equal parts. Through each
point of division, draw a line parallel to BC.
According to Theorem 11, the parallel lines cut off equal segments of equal length along [AC].
Let k be the length of each of these equal segments.
Proof:
⇒ |AY|= sk and |YC| = tk
Q.E.D.
Let ABC be a triangle. If the line l is parallel to BC and cuts [AB] in the ratio s : t then it
also cuts [AC] in the same ratio.
⇒ |AY|: |YC| = sk : tk = s : t
A
B C
X Y
k
Proof: Theorem 12
10 Geometry II](https://image.slidesharecdn.com/chapter10-240101155243-cf494de2/85/Chapter-10-pptx-7-320.jpg)
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ACTIVE MATHS 4 BOOK 2
If two triangles ABC and DEF are similar then their sides are proportional in order.
Given: Similar triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹.
To prove:
|𝐴𝐵|
|𝐷𝐸|
=
|𝐵𝐶|
|𝐸𝐹|
=
|𝐴𝐶|
|𝐷𝐹|
Construction: Assume triangle DEF is smaller
than the triangle ABC.
• Mark a point X on [AB] such that |AX| = |DE|.
• Mark a point Y on [AC] such that |AY| = |DF|. • Draw [XY].
A
B C
D
E F
X Y
Proof: Δ AXY is congruent to Δ DEF…………….SAS
⇒ ∠AXY = ∠𝐴𝐵𝐶
⇒
|𝐴𝐵|
|𝐴𝑋|
=
|𝐴𝐶|
|𝐴𝑌|
…………….Theorem 12
But 𝐴𝑋 = 𝐷𝐸 and 𝐴𝑌 = |𝐷𝐹| …………….Construction
⇒
|𝐴𝐵|
|𝐷𝐸|
=
|𝐴𝐶|
|𝐷𝐹|
⇒ XY // 𝐵𝐶…………….Corresponding angles equal
Similarly
|𝐵𝐶|
|𝐸𝐹|
=
|𝐴𝐵|
|𝐷𝐸|
⇒
|𝐴𝐵|
|𝐷𝐸|
=
|𝐵𝐶|
|𝐸𝐹|
=
|𝐴𝐶|
|𝐷𝐹| Q.E.D.
Proof: Theorem 13
10 Geometry II](https://image.slidesharecdn.com/chapter10-240101155243-cf494de2/85/Chapter-10-pptx-8-320.jpg)
Chapter 10.pptx
- 1. 1 ACTIVE MATHS 4 BOOK 2 Learning Outcomes In this chapter you have learned: Chapter 10 – Geometry II 10 • The following terms related to logic and deductive reasoning: theorem; proof; axiom; corollary; converse; implies; is equivalent to; if and only if; proof by contradiction • Axiom 4. Congruent triangles (SSS, SAS, ASA and RHS) • To prove theorems 11, 12 and 13
- 2. 2 ACTIVE MATHS 4 BOOK 2 Geometry II 10 Axiom, Theorem, Proof, Corollary, Converse and Implies An axiom is a rule or statement that we accept without proof. Example: Axiom 1 (two points axiom): There is exactly one line through any two given points. A theorem is a rule that has been proven by following a certain number of logical steps or by using a previous theorem or axiom that you already know. Example: Theorem 17: A diagonal of a parallelogram bisects the area. A corollary is a statement that follows readily from a previous theorem. Example: Corollary 3: Each angle in a semicircle is a right angle. Play Video The converse of a theorem is the reverse of the theorem. Example: Statement: In an isosceles triangle the angles opposite equal sides are equal. Converse: If two angles are equal in a triangle, then the triangle is isosceles. NOTE: A converse of a statement may not be true. Example: Statement: In a square, opposite sides are equal in length (True). Converse: If opposite sides are equal in length, then it is a square (False). Implies is a term used in a proof when we can write down a fact we have proved from our previous statements. The symbol for implies is ⇒. A proof is a series of logical steps that we use to prove a theorem. Definitions
- 3. 3 ACTIVE MATHS 4 BOOK 2 Definitions In similar or equiangular triangles, all three angles in one triangle have the same measure as the corresponding three angles in the other triangle. 5 10 = 1 2 The corresponding sides of similar triangles are in the same ratio. 9 18 = 1 2 7 14 = 1 2 Theorem 13: If two triangles are similar then their sides are proportional, in order. Play Video NOTE: If a triangle is cut by a line parallel to one of its sides, this line divides the triangle into two similar triangles. 10 Geometry II Proofs: An Introduction • Axiom 4. Congruent triangles (SSS, SAS, ASA and RHS). Congruent triangles Similar triangles
- 4. 4 ACTIVE MATHS 4 BOOK 2 To prove: 2 is irrational. This is an example of proof by contradiction. 2 = 𝑝 𝑞 ⟹ 2 = 𝑝2 𝑞2 (𝑠𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠) ∴ 𝑝2 = 2𝑞2 so 𝑝2 is clearly even. ∴ p = 2m, where m is an integer ∴ 𝑝2 = 4𝑚2 𝑠𝑞𝑢𝑎𝑟𝑖𝑛𝑔 𝑏𝑜𝑡ℎ 𝑠𝑖𝑑𝑒𝑠 . 𝐴𝑠 𝑝2 = 2𝑞2 and 𝑝2 = 4𝑚2 ⟹ 2𝑞2= 4𝑚2 ⟹ 𝑞2 = 2𝑚2 If 𝑝2 is even then 𝑝 must be even ∴ 𝑞2 is clearly even ⟹ 𝑞 is even. If p and q are both even they have 2 as a common factor. This contradicts that they have no common factor other than 1. Thus, our assumption is incorrect and therefore 2 is irrational. Proof: Assume that 2 is rational and can therefore be written in the form of 𝑝 𝑞 where 𝑝 and 𝑞 have no common factors. Proof that 𝟐 is Irrational 10 Geometry II
- 5. 5 ACTIVE MATHS 4 BOOK 2 Useful Terms Geometrical cuts: In our course, we may encounter ‘geometrical cuts’. These are questions that can usually be answered with a series of statements and the reasons behind each statement. These ‘cuts’ could be considered as mini-proofs. Is equivalent to: This means that something has the same value or measure as, or corresponds to, something else. For example, $2 is equivalent to €1.50. If and only if: Ben will go to the cinema if, and only if, his favourite film is on. This means that if his favourite film is on, Ben will go to the cinema and if he is going to the cinema then his favourite film is on. Geometrical example: Two lines are parallel if, and only if, for any transversal, corresponding angles are equal. Proof by contradiction (indirect proof): Occasionally, there are times when we cannot directly prove a statement. However, we can show that the statement cannot be false. This is known as proof by contradiction – we prove that a statement or assumption is true by showing that the statement or assumption being false would imply a contradiction (impossibility). For example, the statement ‘If x > 10 then x > 15’ is not true. We can disprove the statement by letting x = 11. Proofs: Theorems and Terms 10 Geometry II
- 6. 6 ACTIVE MATHS 4 BOOK 2 If three parallel lines cut off equal segments on some transversal line then they cut off equal segments on any other transversal. Given: AD // BE // CF, as in the diagram,with|AB|= |BC|. To prove: |DE|= |EF| Construction: Draw AE’ // DE, cutting EB at E’ and CF at F’. E’ F’ Draw F’B’ // AB, cutting EB at B’ as in the diagram. B’ A’ Statement Reason |B’F’| = |BC| Opposites sides of a parallelogram = |AB| By assumption ∠𝐵𝐴𝐸′ = ∠𝐸′ 𝐹′𝐵′ Alternate angles Proof: ∠𝐴𝐸′ 𝐵 = ∠𝐹′ 𝐸′ 𝐵′ Vertically opposite angles ∴ Δ 𝐴𝐵𝐸′ is congruent to Δ 𝐹′ 𝐵′𝐸′ ASA ∴ |AE’| = |F’E’| But 𝐴𝐸′ = 𝐷𝐸 and 𝐹′𝐸′ = 𝐹𝐸 Opposite sides of a parallelogram ∴ |DE| = |FE| Q.E.D. (Latin phrase quod erat demonstrandum, meaning "which is what had to be proven“). Proof: Theorem 11 10 Geometry II
- 7. 7 ACTIVE MATHS 4 BOOK 2 Given: A Δ 𝐴𝐵𝐶 and a line 𝑋𝑌 parallel to 𝐵𝐶 which cuts 𝐴𝐵 in the ratio 𝑠 ∶ 𝑡. To prove: |AY| : |YC| = s : t. Construction: Divide [AX] into s equal parts and [XB] into t equal parts. Through each point of division, draw a line parallel to BC. According to Theorem 11, the parallel lines cut off equal segments of equal length along [AC]. Let k be the length of each of these equal segments. Proof: ⇒ |AY|= sk and |YC| = tk Q.E.D. Let ABC be a triangle. If the line l is parallel to BC and cuts [AB] in the ratio s : t then it also cuts [AC] in the same ratio. ⇒ |AY|: |YC| = sk : tk = s : t A B C X Y k Proof: Theorem 12 10 Geometry II
- 8. 8 ACTIVE MATHS 4 BOOK 2 If two triangles ABC and DEF are similar then their sides are proportional in order. Given: Similar triangles 𝐴𝐵𝐶 and 𝐷𝐸𝐹. To prove: |𝐴𝐵| |𝐷𝐸| = |𝐵𝐶| |𝐸𝐹| = |𝐴𝐶| |𝐷𝐹| Construction: Assume triangle DEF is smaller than the triangle ABC. • Mark a point X on [AB] such that |AX| = |DE|. • Mark a point Y on [AC] such that |AY| = |DF|. • Draw [XY]. A B C D E F X Y Proof: Δ AXY is congruent to Δ DEF…………….SAS ⇒ ∠AXY = ∠𝐴𝐵𝐶 ⇒ |𝐴𝐵| |𝐴𝑋| = |𝐴𝐶| |𝐴𝑌| …………….Theorem 12 But 𝐴𝑋 = 𝐷𝐸 and 𝐴𝑌 = |𝐷𝐹| …………….Construction ⇒ |𝐴𝐵| |𝐷𝐸| = |𝐴𝐶| |𝐷𝐹| ⇒ XY // 𝐵𝐶…………….Corresponding angles equal Similarly |𝐵𝐶| |𝐸𝐹| = |𝐴𝐵| |𝐷𝐸| ⇒ |𝐴𝐵| |𝐷𝐸| = |𝐵𝐶| |𝐸𝐹| = |𝐴𝐶| |𝐷𝐹| Q.E.D. Proof: Theorem 13 10 Geometry II