Question

prove that if the square of one side of a triangle is equal to the sum of the squares of the other two sides then the angle opposite to the first side is right angle​

Answer 1

Step-by-step explanation:

Given:in ∆ABC,(AC)*=(AB)*+(BC)*

To prove:angle B is 90°

Construction:construct a triangle PQR in which angle Q is 90°and AB=PQandBC=QR.

proof:In ∆PQR we have,

(PR)*=(RQ)*+(PQ)*--------eq(1)

(by p.th as angleQ is90°)

also given that,

(AC)*=(AB)*+(BC)*---eq(2)

from eq(1)and(2),

(PR)*=(BC)*+(AB)*--------eq(3)

(by construction)

from eq(2)and(3),we have

(AC)*=(PR)* (square to square cancel )

AC=PR-----------eq(4)

Now,In ∆ABCand∆PQR,we have

AC=PR{from (eq1)}

QR=BC{by construction}

AB=PQ{by construction}

so, ∆ABCis similar to ∆PQR

(by SSS criteria of congruence)

Therefore,by CPCT,

angleB=angleQ-------eq(5)

but,angleQ=90°-------eq(6)

from eq (5)and eq(6),

angleB=90°

Hence prove.