Question

ABCD is a rhombus and the diagonals intersect at O. Prove that AO^2+OC^2= AD^2+CD^2-1/2BD^2​

Answer 1

Hello, Buddy!!

Refer The Attachment ⤴️

Hence, Proved!!

• $\sf{(AO)²+(OC)² = (AD)²+(CD)²-½(BD)²}$

@MrMonarque♡

Hope It Helps You ✌️

Answer 2

SOLUTION :-

We know that,

→ Diagonals of a rhombus bisect each other.

So, $\sf{OD \perp AC}$

$\sf{\therefore \triangle AOD = \triangle COD = 90 \degree}$

By Pythagoras Theorem,

In △ AOD,

⇒(AO)² = (AD)² - (OD)² -------------❶

In △COD,

⇒ (AO)² + (OC)² = (AD)² + (DC)² - 2(OD)² --❷

$\therefore$ OD = OB = ½BD

From equation ❷,

$\sf \small{→{(AO)}^{2} + {(OC)}^{2} = {(AD)}^{2} + {(DC)}^{2} - 2{[\frac{BD}{2}]}^{2} }$

$\sf \small{→{(AO)}^{2} + {(OC)}^{2} = {(AD)}^{2} + {(DC)}^{2} - { \cancel 2} [\frac{ {(BD)}^{2} }{ \cancel 4}]}$

$\therefore \sf \small{{(AO)}^{2} + {(OC)}^{2} = {(AD)}^{2} + {(DC)}^{2} - \frac{1}{2} {(BD)}^{2} }$

Hence Proved.