Question

In the adjoining figure, ABCDis a rhombus and the diagonals AC and BD intersect each other at O
Prove that ABpower2 +BCpower2+ CDpower2+ DApower2=4 (OApower2+OBpower2)

Answer 1

As diagonals of rhombus intersect each other at right angle
Therefore in right triangle aob
Ab*2=oa*2+ob*2. (1)
Similarly in triangle boc, cod and doa
BC*2=ob*2+oc*2,CD*2=oc*2+od*2 &
Da*2=od*2 +oa*2. (2,3,4)
Add 1,2,3,4
Ab*2+BC*2+CD*2+da*2=2(oa*2+ob*2+oc*2+od*2)
Oc =oa and od=on
Therefore
Ab*2+BC*2+CD*2+da*2=4(oa *2+ob*2)