In a quadrilateral $A B C D$ , $∠ B = 90$ , ${A D}^{2} = {A B}^{2} + {B C}^{2} + {C D}^{2}$ , prove that $∠ A C D = 90^{0}$

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Solution

As $∠ A B C = 90^{\circ}$

So applying Pythagoras theorem in $△ A B C$
$A B^{2} + B C^{2} = A C^{2}$ (1)

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

Substituting (1) in (2)
$A D^{2} = A C^{2} + C D^{2}$

In $△ A C D$ , applying converse of Pythagoras theorem which states
that if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.
Hence $∠ A C D = 90^{\circ}$

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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^{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 C D, ∠ B = 90^{o}

∠ D = 90^{o}

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

∠ A + ∠ D = 90^{0}

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

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