Question

the diagonals ac and bd of a rhombus intersect each other at o. prove that ab2+ bd2+ cd2+ da2= 4( oa2+ ob2)​

Answer 1

In a rhombus all sides are equal and diagonals bisect each at 90°

So,BO=OD and AO=OC ----(1)

So, by Pythagoras theorem,

AB^2 = AO^2 + BO^2 ----(i)

BC^2 = BO^2 + CO^2 ----(ii)

CD^2 = CO^2 + DO^2 ----(iii)

AD^2 = AO^2 + DO^2 ----(iv)

adding (i),(ii),(iii) & (iv)

AB^2 + BC^2 + CD^2 + AD^2 = AO^2 + BO^2 +BO^2 + CO^2 + CO^2 + DO^2 + AO^2 + DO^2

from (1) ---> Replace OC by AO and OD by BO

AB^2 + BC^2 + CD^2 + AD^2 = AO^2 +BO^2 + BO^2 + AO^2 + AO^2 + BO^2 + AO^2 +BO^2

AB^2 + BC^2 + CD^2 + AD^2 = 4AO^2 + 4BO^2

AB^2 + BC ^2 + CD^2 + AD^2 = 4(AO^2 + 4BO^2)

HENCE PROVED

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Answer 2

Step-by-step explanation:

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