State true or false:
In quadrilateral ABCD; $∠ B = ∠ D = 90^{o}$ .
then, $A B^{2} - A D^{2} = C D^{2} - C B^{2}$

True

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False

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Solution

The correct option is A True
Given: Quadrilateral ABCD, $∠ B = ∠ D = 90^{\circ}$
Join AC,
In $△ A C B$ ,
$A C^{2} = A B^{2} + B C^{2}$ (Pythagoras theorem) (I)
In $△ A C D$ ,
$A C^{2} = C D^{2} + A D^{2}$ (By pythagoras theorem) (II)
Equating I and II
$A B^{2} + B C^{2} = C D^{2} + A D^{2}$
$A B^{2} - A D^{2} = C D^{2} - B C^{2}$

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Similar questions

A B C D, ∠ B = 90^{0}

∠ D = 90^{0}

2 A C^{2} - A B^{2} = B C^{2} + C D^{2} + D A^{2}

A B C D ∠ B = 90^{o}

A D^{2} = A B^{2} + B C^{2} + C D^{2}

∠ A C D = 90^{o}

A B C D

∠ A B C = 90^{o}

{A D}^{2} = ({A B}^{2} + {B C}^{2} + {C D}^{2})

∠ A C D = 90^{o}

A B^{2} + C D^{2} = A C^{2} + D A^{2}

A B C D, ∠ B = 90^{o}

∠ D = 90^{o}

2 {A C}^{2} - {A B}^{2} = {B C}^{2} + {C D}^{2} + {D A}^{2}

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