Question

state and prove that Converse of Pythagoras theorem

Answer 1

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Answer 2

Statement:

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides then the angle opposite to the first side is a right angle.

Given:

In Triangle ABC,

$AC^{2} = AB^{2} + BC^{2}$

To Prove That:

Angle ABC is a right angle

Construction:

Construct a triangle PQR such that AB = PQ and BC = QR and Angle Q = 90 degree

Proof:

In Triangle PQR,

Angle Q = 90 Degree (Given)

$PR^{2} = PQ^{2} + QR^{2}$ (By Pythagoras Theorem)

$PR^{2} = AB^{2} + BC^{2}$ (Using PQ = AB and QR = BC by Construction)--Equation 1

but $AC^{2} = AB^{2} + BC^{2} (Given)$--------- Equation 2

Equating Equation 1 and Equation 2,

$PR^{2} = AC^{2}$

PR = AC

Now, In Triangle ABC and Triangle PQR,

AB = PQ (By Construction)

BC = QR (By Construction)

AC = PR (Proved Above)

Triangle ABC is congruent to triangle PQR by SSS

Angle B = Angle Q = 90 Degree

This shows that Triangle ABC is right angled.

Hence Proved