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Question

In a quadrilateral ABCD, B=90°, AD²=AB²+BC²+CD². Prove that ACD=90°.


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Solution

Proving that ACD=90°

Given: In quadrilateral ABCD

B=90°

AD²=AB²+BC²+CD²

Construction: Join Diagonal AC

Consider ABC

ABC is a right angled triangle at vertex B

By Pythagoras' theorem,

AB2+BC2=AC2 ...(i)

It is given that

AD²=AB²+BC²+CD²

From (i)

AD2=AC2+CD2...(ii)

By the converse of Pythagoras' theorem, if the sum of squares of two sides of triangle is equal to the square of the third side, then the triangle is a right angled triangle with the right angle at the common vertex to the two sides.

Consider ADC

As, AD2=AC2+CD2

ADC is a right angled triangle at vertex C

Hence, ACD=90°.

Hence proved.


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