Question

In any quadrilateral ABCD ∠A + ∠D = 90°. Prove that AC² +BD² = AD² + BC².

Answer 1

Given : A quad. ABCD in which, ∠A + ∠D = 90°

To Prove : AC² + BD² = AD² + BC².

Construction : Produce AB and DC intersect  each other at E.

Proof : ∠A + ∠D = 90°

In triangle ADE

∠A + ∠D + ∠E = 180°

⇒ 90° + ∠E = 180°

⇒ ∠E = 180° – 90° = 90°

In right triangle BEC

BC² = BE² + CE²

(By Pythagoras Theorem)

AD² = AF² + DE² …..(ii)

Adding equation (i) and (ii)

BC² + AD² = BE² + AE² + CE² + DE² …..(iii)

In right ∆AEC,

AC² = AE² + CE² …(iv)

In right angled triangle BED

BD² = BE² + DE² ……(v)

From equation (iv) and (v)

AC² + BD² = BE² + AE² + CE² + DE² …..(vi)

From equations (iii) and (iv)

BC² + AD² = AC² + BD²

⇒ AC² + BD² = AD² + BC².