In an isosceles triangle $A B C$ with $A B = A C,$ $B D$ is perpendicular from $B$ to side $A C .$ Prove that $B D^{2} - C D^{2} = 2 C D . A D$

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Solution

Since $△ A D B$ is right-angled at $D,$

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

$\Rightarrow$ $A C^{2} = A D^{2} + B D^{2}$ $(∵ A B = A C)$

$\Rightarrow$ ${(A D + C D)}^{2} = A D^{2} + B D^{2}$

$\Rightarrow$ $A D^{2} + C D^{2} + 2 A D . C D = A D^{2} + B D^{2}$

$\Rightarrow$ $B D^{2} - C D^{2} = 2 A D . C D$

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