Question

prove that the sum of the squares of
the sides of a rhombus is equal to sum
of the
Squares
of its diagonals​

Answer 1

Answer:

□ABCD is a rhombus with O as point of intersection of diagonals.

In ΔAOB,∠AOB=90 0

(since diagonals are perpendicular in rhombus).

By Pythagoras theorem,AB 2

=AO 2+OB 2

Similarly,BC 2

=OC 2+OB 2,DC 2

=OD 2 +OC 2DA 2

=DO 2+OA 2 AB 2+BC 2+CD 2+DA 2

=2(OA 2 +OB 2+OC 2+OD 2

=4(AO 2+DO 2 )

Rhombus diagonal biset each other,

AO=OC,DO=OB

AC=AO+OC

AC 2

=OA 2+OC 2+2AO.OC=4AO 2

Similarly,DB 2

=4OD 2

∴AC 2+DB 2

=4(AO 2+DO 2 )AB 2+BC 2 +CD 2+DA 2

=AC 2+DB