In $△ A B C, A B = A C, B D ⊥ A C$ prove that $B D^{2} - C D^{2} = 2 A D . C D$

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Solution

Given:- A $△ A B C$ in which $A B = A C$ and $B D ⊥ A C$ .
To prove:- ${B D}^{2} - {C D}^{2} = 2 A D \cdot C D$
Proof:-
In right triangle $A B D$ ,
Using pythagoras theorem,
${A B}^{2} = {A D}^{2} + {B D}^{2}$
${A C}^{2} = {A D}^{2} + {B D}^{2} (∵ A B = A C)$
${(A D + D C)}^{2} = {A D}^{2} + {B D}^{2}$
${A D}^{2} + {C D}^{2} + 2 A D \cdot C D = {A D}^{2} + {B D}^{2}$
$\Rightarrow {B D}^{2} - {C D}^{2} = 2 A D \cdot C D$
Hence proved.

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