In triangle ABC, AB=AC and BD is perpendicular to AC prove that $B C^{2} + C D^{2} = 2 A C \times C D$

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Solution

Given: BD is perpendicular to AC and AB = AC
To prove: $B C^{2} = 2 A C . C D$
Proof: In $Δ B C D$ by pythagoras theorem, we have
$\Rightarrow B C^{2} = B D^{2} + C D^{2} \dots \dots (1)$
Again, in $Δ A B D$ by pythagoras theorem
$A B^{2} = B D^{2} + A D^{2} \Rightarrow B D^{2} = A B^{2} - A D^{2}$
On putting value of $B D^{2}$ in (1), we get
$B C^{2} = A B^{2} - A D^{2} + C D^{2} \Rightarrow B C^{2} = A B^{2} - (A C - C D)^{2} + C D^{2} \Rightarrow B C^{2} = A B^{2} - A C^{2} - C D^{2} + 2 A C . C D + C D^{2} \Rightarrow B C^{2} = 2 A C . C D$ [Given, $A B = A C \Rightarrow A B^{2} = A C^{2}$ ]
[Hence proved]
and $B C^{2} = B D^{2} + C D^{2}$

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