1. Prove that the sum of the squares of the sides of a rhombus
is equal to the sum of the squares of its diagonals.
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Given :- A rhombus ABCD with diagonals AC and BD intersecting at point O.
To proove : - AB2 + BC2 + CD2 + AD2 = AC2 + BD2
Proof :- Since we know that diagonals of rhombus intersect each other at 900 .
Therefore, ang. AOB = ang. BOC = ang. COD = ang. AOD = 900
By Pythagores Theorem,
in triangles AOB, BOC, COD and AOD, we will get :-
AB2 = AO2 + BO2 ---------(1)
BC2 = BO2 + CO2 --------(2)
CD2 = CO2 + DO2 -------(3)
AD2 = AO2 + DO2 ---------(4)
On adding (1),(2),(3) and (4),
AB2 + BC2 + CD2 + AD2 = 2 (AO2 + BO2 + CO2 +DO2)
= 2 ( AC2 / 2 + BD2 /2) (DIAGONALS BISECT EACH OTHER.
AO = CO = AC / 2 BO = DO = BD / 2 )
= AC2 + BD2
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