Question

State and prove converse of Pythagoras theorem from similar triangles.​

Answer 1

Question : -

State and prove the converse of Pythagoras theorem !

ANSWER

Statement : -

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

Given : -

In ∆ABC,

AC² = AB² + BC²

Required to prove : -

• ∠B = 90°

Construction : -

Construction an imaginary right angle ∆PQR such that ∠Q = 90° and AB = PQ, BC = QR.

Proof : -

In ∆ABC,

It is given that;

AC² = AB² + BC² ..... (1)

Now,

In ∆PQR,∠Q = 90°

Using the Pythagoras theorem;

(hypotenuse)² = (side)² + (side)²

This implies;

PR² = PQ² + QR²

Since,

• AB = PQ
• BC = QR

[ By construction ]

So,

PR² = (AB)² + (BC)²

PR² = AB² + BC²

From equation - 1

PR² = AC²

Taking square root on both sides

√(PR²) = √(AC²)

★ PR = AC

Now,

Consider ∆ABC & ∆PQR

In ∆ABC & ∆PQR

AB = PQ [ By construction ]

BC = QR [ " " ]

AC = PR [ Proved above ]

Using the SSS congruency rule !

∆ABC ≅ ∆PQR

Here,

∠B = ∠Q [ corresponding parts of congruent triangles ]

This implies;

∠B = 90°

Hence Proved ! ㋡