Question

In a quadrilateral ABCD, angle A + angle D = 90. Prove that AC2 + BD2 = AD2 + BC2 [Hint: Produce AB and DC to meet at E]

Answer 1

Hope, this clear your doubt,

Answer 2

Answer:

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