In a quadrilateral $A B C D ∠ B = 90^{o}$ .If $A D^{2} = A B^{2} + B C^{2} + C D^{2}$ , prove that $∠ A C D = 90^{o}$ .

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Solution

In $Δ A B C$ ,

$A B^{2} + B C^{2} = A C^{2}$ (i)

$A D^{2} = A B^{2} + B C^{2} + C D^{2}$ (given)
$A D^{2} = A C^{2} + C D^{2}$ (from (i))

This condition (Pythagoras theorem) satisfies only when $Δ A C D$ is a right angled triangle with $∠ A C D = 90^{0}$

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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 C D, ∠ B = 90^{o}

∠ D = 90^{o}

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

In a quadrilateral ABCD, $∠ B = 90^{\circ}, A D^{2} = A B^{2} + B C^{2} + C D^{2}$ , prove that $∠ A C D = 90^{\circ}$

A B C D

∠ B = 90

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

∠ A C D = 90^{0}

∠ B = ∠ D = 90^{o}

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

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