In the quadrilateral $A B C D$ , $∠ A B C = 90^{o}$ and ${A D}^{2} = ({A B}^{2} + {B C}^{2} + {C D}^{2})$ . Prove that : $∠ A C D = 90^{o}$

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Solution

In $△ A B C$ , $∠ A B C = 90^{o}$ ............ [Given]
$∴ {A C}^{2} = {A B}^{2} + {B C}^{2}$ .......... [By Pythagoras theorem]
But ${A D}^{2} = ({A B}^{2} + {B C}^{2}) + {C D}^{2}$ [ $∵$ Data]
$∴ {A D}^{2} = {A C}^{2} + {C D}^{2}$ ........... [Given ${A B}^{2} + {B C}^{2} = {A C}^{2}$ ]
$∴ ∠ A C D = 90^{o}$ ......... [By converse of Pythagoras theorem]

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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 $∠ A C D = 90^{\circ}$

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