Question

Prove that the sum of squares on the sides of a rhombus is equal to sum of squares on its duagnals

Answer 1

Answer:Let the rhombus be ABCD where AB//CD. AC and BD are diagonals and they meet at O

In,ΔDOC,

DC^2=OD^2+OC^2(diagonals bisect at 90°-pythagorus theorem.

DC^2=(BD/2)^2+(AC/2)^2(diagonals bisect each other)

DC^2=(DB^2+DC^2)/4

4DC^2=DB^2+DC^2

AB^2+BC^2+DC^2+AC^2=DB^2+AC^2(all the sides are equal)

Hence, proved.