In the given figure, $A D ⊥ B C$ , Prove that ${A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}$

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Solution

In $△ A D C$ , $∠ A D C = 90^{o}$
$∴ {A C}^{2} = {A D}^{2} + {C D}^{2}$ (By Pythagoras theorem) .......... $(1)$
In $△ D B A$ , $∠ A D B = 90^{o}$
$∴ {A B}^{2} = {A D}^{2} + {B D}^{2}$ (By Pythagoras theorem) .......... $(2)$
Subtracting $(1)$ from $(2)$ , we get
${A B}^{2} - {A C}^{2} = {A D}^{2} + {B D}^{2} - {A D}^{2} - {C D}^{2}$
$∴ {A B}^{2} - {A C}^{2} = {B D}^{2} - {C D}^{2}$
$∴ {A B}^{2} + {C D}^{2} = {B D}^{2} + {A C}^{2}$

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