Question

10. In a quadrilateral ABCD, ZB = 90° and
AD^2 = AB^2 + BC^2 + CD^2. Prove that LACD = 90°​

Answer 1

Answer:

✍️✍️Question:

In a quadrilateral ABCD, ZB = 90° and

AD²= AB² + BC² + CD² . Prove that LACD = 90°

✍️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.✍️✍️✍️

Answer 2

Answer:

Question:

In a quadrilateral ABCD, ZB = 90° and

AD²= AB² + BC² + CD² . Prove that LACD = 90°

Solution:

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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