Question

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

Answer 1

SOLUTION :  

Given :  ∠B = 90° and AD² = AB² + BC² +CD²

AD² - CD² = AB² + BC² …………..(1)

In ∆ ABC,∠B = 90°

AC² = AB² + BC² ……………….(2)

From eq 1 & 2,

AD² - CD² = AC²

AC² + CD² = AD²

By CONVERSE OF PYTHAGORAS THEOREM, In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite to the side is a right angle.

Hence, ∠ACD = 90°

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Answer 2

Given : A quadrilateral ABCD in which

           ∠B = 90° and

           $AD^{2} = AB^{2} + BC^{2} + CD^{2}$

To prove : ∠ACD = 90°

Construction : Join AC

Proof : In ΔABC, ∠B = 90°.

∴ $AC^{2} = AB^{2} + BC^{2}$            ......(i) [by Pythaogoras' theorem]

Now, $AD^{2} = AB^{2} + BC^{2} + CD^{2}$               (given)

⇒ $AD^{2} = AC^{2} + CD^{2}$                     [using (i)]

Thus, in ΔACD, we have : $AD^{2} = AC^{2} + CD^{2}$

Hence, ∠ACD = 90°  [by the converse of Pythagoras' theorem]