Question

In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC² - AB² = BC² + CD² + DA²

Answer 1

In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.

In ΔABC, using Pythagoras theorem,
AC² = AB² + BC²
⇒ AB² = AC² - BC²                  ....(i)

In ΔADC, using Pythagoras theorem,
AC² = AD² + DC²                   ....(ii)

LHS = 2AC² - AB²
= 2AC² - ( AC² - BC² )           .....[ From(i) ]
= 2AC² - AC² + BC²
= AC² + BC²
= AD² + DC² + BC²            ....[ From(ii) ]
= RHS