Question

In a quadrilateral, ΔBCD, ∠B = 90°. If AD² = AB² + BC² + CD², prove that ∠ACD = 90°.

Answer 1

Given: ABCD is a quadrilateral and ∠B = 90°

and AD²= AB² + BC² + CD²

To Prove: ∠ACD = 90°

In right triangle ∆ABC, using Pythagoras theorem, we have

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AB)² + (BC)² = (AC)² …(i)

Given: AD²= AB² + BC² + CD²

⇒ AD²= AC² + CD² [from (i)]

In ∆ACD

AD²= AC² + CD²

∴ ∠ACD = 90° [converse of Pythagoras theorem]

Hence Proved