Question

In a quadrilateral ABCD, angle A + angle D = 90 dregree, prove that AC^2 + BD^2 = AD^2 + BC^2.​

Answer 1

Answer:

We have, A + D = 90

In APD, by angle sum property,

A + D + P = 180

90 + P = 180

P = 180- 90 = 90

In APC, by Pythagoras theorem,

AC² = AP² + PC²..(1)

In BPD, by Pythagoras theorem,

BD2= BP²+ DP²...(2)

Adding equations (1) and (2),

AC²+ BD² = AP² + PC² + BP² + DP²

AC²+ BD² = (AP² + DP²) + (PC² + BP²)

AC²+ BD² = AD² + BC²

Step-by-step explanation:

dbb9f1bfd33b2d861fc6546f04dbf546.docx