Question

Diagonals of rhombus ABCD intersect each other at point O.

Prove that: OA² + OC² = 2AD² - `"BD"^2/2`

Answer 1

Diagonals of the rhombus are perpendicular to each other.

In quadrilateral ABCD, ∠AOD = ∠COD = 90°.
So, ΔAOD and ΔCOD are right-angled triangles.

In ΔAOD using Pythagoras theorem,
AD² = OA² + OD²
⇒ OA² = AD² - OD²                 ....(i)

In ΔCOD using Pythagoras theorem,
CD² = OC² + OD²
⇒ OC² = CD² - OD²               ....(ii)

LHS = OA² + OC²
= AD² - OD² + CD² - OD²      ...[ From(i) and (ii) ]
= AD² + CD²  - 2OD²

= AD² + AD² - 2`("BD"/2)^2` ...[ AD = CD and OD = `"BD"/2`]

= 2AD² - `("BD")^2/2`

= RHS.