ABC above is a right triangle. The sides BC and AC are also known as - PowerPoint PPT Presentation

D AY 77 – P ROOF OF THE P YTHAGOREAN THEOREM USING SIMILARITY

I NTRODUCTION The Pythagorean theorem, named in honor of the Greek philosopher, Pythagoras of Samos, is one of the fundamental theorems in geometry. It is also referred to as the Pythagoras’ theorem. It explores the relationship between the squares of the lengths of the three sides of a right triangle. It is possible to prove this theorem using the concept of similar triangles though it can be proved by other ways, including algebra. In this lesson, we will learn how to prove this theorem based on the concept of similarity.

V OCABULARY 1. Hypotenuse The side of a right triangle that is opposite the right angle. 2. Right triangle A triangle that has a right angle as one of its interior angles.

P YTHAGOREAN THEOREM The Pythagorean theorem states that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. A 𝑐 𝑑 C B 𝑏

ΔABC above is a right triangle. The sides BC and AC are also known as the legs of the right triangle while the side AB is known as the hypotenuse . According to the Pythagorean theorem, we have: AC 2 +BC 2 = AB 2 𝑐 2 + 𝑏 2 = 𝑑 2

T HE C ONVERSE OF P YTHAGORAS ’ THEOREM The converse of the Pythagoras’ theorem is equally important: It states that if the sum of the squares of the lengths of two sides of a triangle equals the square of the length of the hypotenuse, then the triangle is a right triangle.

P ROOF OF THE P YTHAGOREAN THEOREM BASED ON SIMILARITY The proof of the theorem is based on the proportionality of the sides of similar triangles and the altitudes of these triangles. We will base our proof on AA similarity criterion for similar triangles, that is, if two corresponding angles of two triangles are congruent, then the two triangles are similar. We have to recall that in similar triangles corresponding angles are congruent and the corresponding sides are proportional.

Consider ΔPQR shown below. P Q R We have to show that PQ 2 +QR 2 = PR 2

We will also use another theorem without proof to prove the Pythagoras theorem. The theorem states that: If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse, then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

A perpendicular is drawn from point Q to meet the hypotenuse, PR at S . QS becomes the altitude of ΔPQR . P S Q R This leads to formation of two sub-triangles inside ΔPQR , namely ΔQSR and ΔPSQ

Let ∠PRQ = θ , it follows that ∠RQS = 90 − θ , ∠PQS = θ and ∠QPS = 90 − θ . P S (90 − 𝜄) 𝜄 (90 − 𝜄) 𝜄 Q R

ΔPQR has ∠P and ∠R ΔPSQ has ∠P ΔQSR has ∠R All the triangles have a right angle. This shows that all the triangles are similar by the AA similarity criterion for similar triangles.

ΔQSR and ΔPSQ are shown below, note that the triangles have been named using the correct correspondence of the sides with respect to ΔPQR . Q P S Q P S S R Q R

QR on ΔPQR corresponds to SR on ΔQSR hence the ratio becomes: QR SR Similarly, PR on ΔPQR corresponds to QR on ΔQSR hence the ratio becomes: PR QR

The lengths of corresponding sides are in the same ratio in similar triangles, hence QR SR = PR QR After cross multiplication, we have: (QR) 2 = PR × SR

PQ on ΔPQR corresponds to PS on ΔPSQ hence the ratio becomes: PQ PS Similarly, PR on ΔPQR corresponds to PQ on ΔPSQ hence the ratio becomes: PR PQ

The lengths of corresponding sides are in the same ratio in similar triangles, hence PQ PS = PR PQ After cross multiplication, we have: (PQ) 2 = PR × PS

Our aim is to show that PQ 2 +QR 2 = PR 2 . Substituting the expressions for PQ 2 and QR 2 we have: PQ 2 +QR 2 = PR × PS + PR × SR PR is a common factor hence: PQ 2 +QR 2 = PR PS + SR but from ΔPQR , 𝐐𝐓 + 𝐓𝐒 = 𝐐𝐒 , hence 𝐐𝐑 𝟑 +𝐑𝐒 𝟑 = PR PS + SR = 𝐐𝐒 𝟑

Therefore, 𝐐𝐑 𝟑 +𝐑𝐒 𝟑 = 𝐐𝐒 𝟑 , which proves the Pythagorean theorem based on similarity of triangles.

Example Use right ΔABC shown below to answer the questions that follow. BC = a, AC = b , AB = c , AD = d, BD = e and CD = f . C 𝑐 𝑏 𝑔 𝑒 𝑓 A B D 𝑑

(a) Write an algebraic expression for 𝑑 in terms of 𝑒 and 𝑓 . (b) Identify two triangles similar to ΔABC and label their vertices in correct correspondence with respect to ΔABC . (c) Complete the following proportionality statements based on the similar triangles you have identified in (c) above and hence fill in the blank spaces. 𝑐 𝑒 = 𝑐 hence 𝑐 2 = __ × 𝑒 (i) 𝑏 𝑓 = 𝑏 hence 𝑏 2 = __ × 𝑓 (ii) (d) Show that 𝑏 2 + 𝑐 2 = 𝑑 2 using substitution of the equations you have identified in (c) above.

Solution (a) From the figure AB = c = AD + BD = d + e , therefore, 𝒅 = 𝒆 + 𝒇 (b) We name the triangles using the correct correspondence. These triangles are ΔACD and ΔCBD (c) From the figure above side 𝑐 on ΔABC corresponds to side 𝑒 on ΔACD and side 𝑑 on ΔABC corresponds to side 𝑐 ΔACD hence 𝑐 𝑐 hence 𝑐 2 = 𝑑 × 𝑒 𝑑 (i) 𝑒 =

From the figure above side 𝑏 on ΔABC corresponds to side 𝑓 on ΔCBD and side 𝑑 on ΔABC corresponds to side 𝑏 ΔCBD hence 𝑏 𝑏 hence 𝑏 2 = 𝑑 × 𝑓 𝑑 (ii) 𝑓 = (d) We are required to show that 𝑏 2 + 𝑐 2 = 𝑑 2 This is an identity, hence 𝑏 2 + 𝑐 2 = 𝑑𝑓 + 𝑑𝑒 𝑏 2 = 𝑑𝑓 and 𝑐 2 = 𝑑𝑒 𝑏 2 + 𝑐 2 = 𝑑(𝑓 + 𝑒) but 𝑑 = 𝑓 + 𝑒 Thus, 𝑏 2 + 𝑐 2 = 𝑑 2

HOMEWORK Use right ΔKLM to answer the questions that follow. LN ⊥ KM , KM = 𝑚 , LM = 𝑙 , KL = 𝑛 , KN = 𝑜 and 𝑂𝑁 = 𝑞 K N 𝑜 𝑚 𝑞 𝑛 M L 𝑙

(a) Express 𝑚 in terms of 𝑜 and 𝑞 (b) Use ΔKLM and ΔKNL to complete the proportionality equation below: 𝑛 𝑜 = 𝑛 hence 𝑛 2 = ___ × 𝑜 (c) Use ΔKLM and ΔLNL to complete the proportionality equation below: 𝑙 𝑞 = 𝑙 hence 𝑙 2 = ___ × 𝑞 (d) Hence prove the identity 𝑛 2 + 𝑙 2 = 𝑚 2

A NSWERS TO HOMEWORK (a) 𝑚 = 𝑜 + 𝑞 𝑛 𝑛 hence 𝑛 2 = 𝑚 × 𝑜 𝑚 (b) 𝑜 = 𝑙 𝑙 hence 𝑙 2 = 𝑚 × 𝑞 𝑚 (c) 𝑞 = (d) 𝑛 2 + 𝑙 2 = 𝑚𝑜 + 𝑚𝑞 = 𝑚(𝑜 + 𝑞) but 𝑜 + 𝑞 = 𝑚 Hence 𝑛 2 + 𝑙 2 = 𝑚 2

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