Question

If the square on one side of a triangle is equal to the sum of the squares on the other two sides, prove that the triangle is right-angled

Answer 1

Let, ABC is a triangle

Given: A ΔABC such that AC²=AB²+BC²

To prove: the ΔABC is rt.<d, at B such that, <ABC=1 rt.<

Construction: Draw BD⊥AB, making BD=BC, join AD.

Proof: AC²=AB²+BC²(Given)

                 = AB²+BD²

                 = AD²

∴ AC=AD

In Δs. ABC and ABD, AB is common

AC=AD(Proved)

BC=BD(const.)

Δ ABC ≅ Δ ABD (S.S.S. axiom)

∴ ∠ABC = ∠ ABD = 1 rt.∠ (∠ABC =1 rt. ∠(const.))

Hence, ΔABC is a right angled triangle (is proved).

Answer 2

Answer:

Hence Proved.

There also exists a Converse of the Pythagoras theorem that states, “If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle”.