prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagnols ?
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refer to the attachment...
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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)
AC²+BD²= 4AB²
AC²+BD²= AB²+AB²+AB²+AB²
AB²+BC²+CD²+DA²= AC²+BD²
AB=BC=CD=DA
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