Question

In a quad. ABCD ANGLE A + ANGLE D =90 DEGREE. THEN PROVE THAT AC^2 + BD^2=BC^2 + AD^2....​

Answer 1

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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°

Answer 2

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In triangle AEC ,

AE^2 + EC^2 = AC^2 --------------- [1]

In triangle BED ,

BE^2 + ED^2 =BD^2 --------------- [2]

Adding [1] and [2] ,

AC^2 + BD^2 = AE^2 + EC^2 + BE^2 + ED^2 --------------------- [3]

In triangle AED ,

AE^2 + ED^2 = AD^2 ---------------- [4]

In triangle BEC ,

BE^2 + EC^2 = BC^2 ---------------- [5]

Adding [4] and [5] ,

AD^2 + BC^2 = AE^2 + ED^2 +BE^2 + EC^2 = AE^2 + EC^2 + BE^2 + ED^2 ------------------------- [6]

From [3] and [6] ,

AC^2 +BD^2 = AD^2 + BC^2

Hence proved.