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In a quadrilateral ABCD, \angle A + \angle D = 90^{o}. Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2}

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Given : ABCD is a quadrilateral, \angle A+\angle D=90^{o}

To prove :- AC^{2} + BD^{2} = AD^{2} + BC^{2}

Proof : In \Delta ADE

\angle A+\angle D=90^{o} \; \; \; \; \; ....(1)  (given)

for find \angle E use sum of the angle of a triangle is equal to 180^0

\angle A+\angle D+\angle E=180^{o}

90+\angle E=180^{o}

\angle E=180^{o} -90^{o}

\angle E=90^{o}

In \Delta ADE use the Pythagoras theorem we get

AD^{2}=AE^{2}+DE^{2}\; \; \; \; \; \; \; \; ....(2)

In \Delta BEC use the Pythagoras theorem we get

BC^{2}=BE^{2}+EC^{2} ....(3)

Add equation (2) and (3) we get

AD^{2}+BC^{2}=AE^{2}+DE^{2}+BE^{2}+CE^{2} .....(4)

In \Delta ACE use the Pythagoras theorem we get

AC^{2}=AE^{2}+CE^{2}....(5)

In \Delta EBD use the Pythagoras theorem we get

BD^{2}=BE^{2}+DE^{2}\; \; \; \; \; \; \; \; \; \; \; \; ....(6)

Now add equation (5) and (6) we get

AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}....(7)

From equation (4) and (7) we get

AC^{2} + BD^{2} = AD^{2} + BC^{2}

Hence proved.

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