Question

prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagnols ?​

Answer 1

refer to the attachment...

Answer 2

Given:-

• ABCD is a rhombus
• AB=BC= DC=DA
• AC perpendicular to BD
• AO=OC=BO=OD

RTP:-

• AB²+BC²CD²+DA²=AC²+BD²

Proof:-

In triangle AOB, Angle AOB = 90°

So, OA²+OB²=AB²(By Pythagoras theorem)

$( \frac{AC}{2}² ) + ( \frac{BD²}{2} ) = AB²$

$\frac{AC² + BD²}{4} = AB²$

AC²+BD²= 4AB²

AC²+BD²= AB²+AB²+AB²+AB²

AB²+BC²+CD²+DA²= AC²+BD²

AB=BC=CD=DA