SLIDE 1
DAY 77 β PROOF OF THE PYTHAGOREAN THEOREM USING
SIMILARITY
SLIDE 2 INTRODUCTION
The Pythagorean theorem, named in honor of the Greek philosopher, Pythagoras of Samos, is one
- f the fundamental theorems in geometry. It is also
referred to as the Pythagorasβ theorem. It explores the relationship between the squares of the lengths
- f 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.
SLIDE 3 VOCABULARY
The side of a right triangle that is opposite the right angle.
A triangle that has a right angle as one of its interior angles.
SLIDE 4
PYTHAGOREAN 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 B C π π π
SLIDE 5
ΞABC 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: AC2+BC2 = AB2 π2 + π2 = π2
SLIDE 6 THE CONVERSE OF PYTHAGORASβ
THEOREM The converse of the Pythagorasβ 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
- f the length of the hypotenuse, then the triangle is
a right triangle.
SLIDE 7
PROOF OF THE PYTHAGOREAN 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.
SLIDE 8
Consider ΞPQR shown below. We have to show that PQ2+QR2 = PR2
P R Q
SLIDE 9 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
SLIDE 10
A perpendicular is drawn from point Q to meet the hypotenuse, PR at S. QS becomes the altitude of ΞPQR. This leads to formation of two sub-triangles inside ΞPQR, namely ΞQSR and ΞPSQ
P R Q S
SLIDE 11
Let β PRQ = ΞΈ, it follows that β RQS = 90 β ΞΈ , β PQS = ΞΈ and β QPS = 90 β ΞΈ .
P R Q S π (90 β π) π (90 β π)
SLIDE 12
ΞPQR has β P and β R ΞPSQ has β P ΞQSR has β R All the triangles have a right angle. This shows that all the triangles are similar by the AA similarity criterion for similar triangles.
SLIDE 13
ΞQSR and ΞPSQ are shown below, note that the triangles have been named using the correct correspondence of the sides with respect to ΞPQR.
Q S R P S Q P R Q S
SLIDE 14
QR on ΞPQR corresponds to SR on ΞQSR hence the ratio becomes:
QR SR
Similarly, PR on ΞPQR corresponds to QR on ΞQSR hence the ratio becomes:
PR QR
SLIDE 15
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 Γ SR
SLIDE 16
PQ on ΞPQR corresponds to PS on ΞPSQ hence the ratio becomes:
PQ PS
Similarly, PR on ΞPQR corresponds to PQ on ΞPSQ hence the ratio becomes:
PR PQ
SLIDE 17
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 Γ PS
SLIDE 18
Our aim is to show that PQ2+QR2 = PR2. Substituting the expressions for PQ2 and QR2we have: PQ2+QR2 = PR Γ PS + PR Γ SR PR is a common factor hence: PQ2+QR2 = PR PS + SR but from ΞPQR , ππ + ππ = ππ, hence πππ+πππ = PR PS + SR = πππ
SLIDE 19
Therefore, πππ+πππ = πππ, which proves the Pythagorean theorem based on similarity of triangles.
SLIDE 20
Example Use right ΞABC shown below to answer the questions that follow. BC = a, AC = b, AB = c, AD = d, BD = e and CD = f.
A B C π π π D π π π
SLIDE 21
(a) Write an algebraic expression for π in terms of π and π. (b) Identify two triangles similar to ΞABC and label their vertices in correct correspondence with respect to ΞABC. (c) Complete the following proportionality statements based on the similar triangles you have identified in (c) above and hence fill in the blank spaces. (i)
π π = π hence π2 = __ Γ π
(ii)
π π = π hence π2 = __ Γ π
(d) Show that π2 + π2 = π2 using substitution of the equations you have identified in (c) above.
SLIDE 22 Solution (a) From the figure AB = c = AD + BD = d + e, therefore, π
= π + π (b) We name the triangles using the correct
- correspondence. These triangles are ΞACD and
ΞCBD (c) From the figure above side π on ΞABC corresponds to side π on ΞACD and side π on ΞABC corresponds to side π ΞACD hence (i)
π π = π π hence π2 = π Γ π
SLIDE 23
From the figure above side π on ΞABC corresponds to side π on ΞCBD and side π on ΞABC corresponds to side π ΞCBD hence (ii)
π π = π π hence π2 = π Γ π
(d) We are required to show that π2 + π2 = π2 This is an identity, hence π2 + π2 = ππ + ππ π2 = ππ and π2 = ππ π2 + π2 = π(π + π) but π = π + π Thus, π2 + π2 = π2
SLIDE 24 HOMEWORK Use right ΞKLM to answer the questions that
- follow. LN β₯ KM, KM = π, LM = π, KL = π, KN = π
and ππ = π
K L M N π π π π π
SLIDE 25
(a) Express π in terms of π and π (b) Use ΞKLM and ΞKNL to complete the proportionality equation below:
π π = π hence π2 = ___ Γ π
(c) Use ΞKLM and ΞLNL to complete the proportionality equation below:
π π = π hence π2 = ___ Γ π
(d) Hence prove the identity π2 + π2 = π2
SLIDE 26
ANSWERS TO HOMEWORK
(a) π = π + π (b)
π π = π π hence π2 = π Γ π
(c)
π π = π π hence π2 = π Γ π
(d) π2 + π2 = ππ + ππ = π(π + π) but π + π = π Hence π2 + π2 = π2
SLIDE 27
THE END
