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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.

Posted by

infoexpert23

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