Diagonals of rhombus ABCD intersect each other at point O. Prove that :

OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$

By formula,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + Base²

In right angle triangle AOD,

By pythagoras theorem,

⇒ AD² = OA² + OD²

⇒ OA² = AD² - OD² ……….(1)

In right angle triangle COD,

By pythagoras theorem,

⇒ CD² = OC² + OD²

⇒ OC² = CD² - OD² ……….(2)

To prove :

OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$

Solving L.H.S. of the above equation, we get :

⇒ OA² + OC² = AD² - OD² + CD² - OD²

⇒ OA² + OC² = AD² + CD² - 2OD² ………(3)

Since, all sides of a rhombus are equal and diagonals of rhombus bisect each other,

∴ CD = AD and OD = $\dfrac{BD}{2}$

Substituting value of CD and OD in equation (3), we get :

⇒ OA² + OC² = AD² + AD² - $2\Big(\dfrac{BD}{2}\Big)^2$

⇒ OA² + OC² = 2AD² - $2 \times \dfrac{BD^2}{4}$

⇒ OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$ = R.H.S.

Hence, proved that OA² + OC² = 2AD² - $\dfrac{BD^2}{2}$.