Question

In a quadrilateral ABCD, angle B = 90° = angle D. Prove that
2AC2 – BC2 = AB2 + AD2 + DC2.​

Answer 1

GIVEN: A quadrilateral ABCD , angle B = 90° And AD² = AB² + BC² + CD²

TO FIND: angle ACD

AC² = AB² + BC² ( by Pythagoras theorem)… … ..(1)

And AD² = (AB² + BC²) +CD² ( given)

=> AD² = AC² + CD² ( by 1)

So, in triangle ACD, angle opposite to AD = 90° ( by Converse of Pythagoras theorem)

=> <ACD = 90°

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Answer 2

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