Question

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

Answer 1

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

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

To Prove: ∠ACD = 90°

PROOF:
AD² = AB²  + BC² + CD²
AD² -  CD² = AB²  + BC² ……………(1)

In right ∆ABC, ∠B  =90°,
AC² = AB² + BC²……………….(2)
[By Pythagoras theorem]

From eq 1 & 2
AC² = AD² -  CD²
AC² + CD² = AD²

Therefore , ∠ACD = 90°
[By converse of Pythagoras theorem]

Hence, proved.

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Answer 2

Hey...!!!!

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Given: ABCD is a quadrilateral, ∠B = 90° and AD2 = AB2 + BC2 + CD2

To prove: ∠ACD = 90°

Proof: In right ∆ABC,

AC2 = AB2 + BC2 … (1)

Given, AD2 = AB2 + BC2 + CD2

⇒ AD2 = AC2 + CD2 (Using (1)

In ∆ACD,

AD2 = AC2 + CD2

∴ ∠ACD = 90° (Converse of Pythagoras theorem)
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