In the quadrilateral ABCD, angle A + angel D=90degree. Prove that AC square + BD square = AD square + BD square. ?

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In the quadrilateral ABCD, angle A + angel D=90degree. Prove that AC square + BD square = AD square + BD square. ?

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In the quadrilateral ABCD, angle A + angel D=90degree. Prove that AC s...

Step-by-step explanation:

We have, ∠A + ∠D = 90�

In ΔAPD, by angle sum property,

∠A + ∠D + ∠P = 180�

 90� + P = 180�

 ∠P = 180� – 90� = 90�

In ΔAPC, by Pythagoras theorem,

AC2 = AP2 + PC2 ....(1)

In ΔBPD, by Pythagoras theorem,

BD2 = BP2 + DP2 ....(2)

Adding equations (1) and (2),

AC2 + BD2 = AP2 + PC2 + BP2 + DP2

 AC2 + BD2 = (AP2 + DP2) + (PC2 + BP2) = AD2 + BC2

Hence proved

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In the quadrilateral ABCD, angle A + angel D=90degree. Prove that AC square + BD square = AD square + BD square. ?

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