Prove that the sum of the squares of
the sides of a shombus is equal to the
Sum of the squares of its diagonals.
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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
2
DA
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
2
Hence Proved.
solution
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