Question

1) Let us Prove that, the sum of squares drawn on the sides of a rhombus is equal to the sum of squares drawn on two diagonals .​

Answer 1

Answer:

Given :-

• ABCD is a rhombus.

To Prove :-

• AB²+ BC²+ CD²+ DA² = AC²+ BD²

Proof :- We know that diagonals of a rhombus bisected each other at right angle.

i.e., AC $\perp$ BD and AO = OC and BO = OD

So, in the right angled triangle ∆AOB, ∆BOC, ∆COD and ∆DOA.

Hence, we get,

➛ AB²= OA²+ OB²,

➛ BC² = OB²+ OC²,

➛ CD²= OC²+ OD²

➛ AD²= OD²+ OA²

Then,

➡ AB²+ BC²+ CD²+ DA²

⇒ OA²+ OB²+OB²+ OC²+ OC²+ OD² + OD²+ OA².

⇒ 2( OA²+ OB²+ OC²+ OD²)

⇒ 2( 2OA²+ 2OB²)

⇒ 4( OA²+ OB²)

Again,

➡ AC² + BD²

↦ (2OA)² + (2OB)²

↦ 4OA² + 4OB²

↦ 4(OA² + OB²)

∴ AB² + BC² + CD² + DA² = AC² + BD² (PROVED)