Converse of Pythagoras Theorem class 10th

Theorem 6.9
In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Proof:- 
Given:- In a △ABC in which 

AC2=AB2+BC2

Prove That:-∠B=90°
Construction:- A △PQR right triangle at Q such that PQ=AB and QR=BC (In figure)
Now, from △PQR , we have:

        PR2=PQ2+QR2   (Pythagoras Theorem as ∠Q=90°)

        PR2=AB2+BC2    (By construction)...(1)

But   AC2=AB2+BC2     (Given)              ...(2)

so,   AC=PR      [from(1) and (2)]...(3)

 Now, in △ABC and △PQR,

                AB=PQ     (by construction)

                BC=QR     (by construction)

                AC=PR      [from (3)]

so,     △ABC≅ △PQR (SSS congrunce)

therefore,∠B =∠C        (CPCT)

But          ∠Q =90°       (by construction)

So,          ∠B = 90°

Hence proved