Question

solve it if u can..no spamming send the pic of Ur copy..

Answer 1

the above image will help you and given below is your answer , sorry as my diagram is not much neat , AM and DM are straight lines.

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firstly , extend the sides AB and CD to join at point M ,

In ∆ AMD

as , angle A + angle B = 90°

by Angle sum property of a triangle ,

angle A + angle B + angle M = 180°

so , 90° + angle M = 180°

angle M = 90°

Now ,In ∆ MAD

AM² + DM² = AD² -- eq-1

In ∆ AMC

AM² + MC² = AC² -- eq-2

In ∆ MBD

BM² + MD² = BD² -- eq-3

In ∆ MBC

BM² + MC² = BC² -- eq-4

putting values of AM and DM from eq -2 and eq -3 into eq -1 , we get ,

AC² - MC² + BD² - BM² = AD²

AC² + BD² = AD² + ( MC² + BM² )

from eq-4 , MC²+BM²=BC² , so ,

AC² + BD² = AD² + BC²

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Hence proved , hoping my answer had surely helped you .