In ∆ABC, given that AB = AC and BD ⊥ AC

In $\triangle A B C$, given that $A B=A C$ and $B D \perp A C$. Prove that $B C^{2}=2 A C . C D$


Since $\triangle A D B$ is right triangle right angled at $D$

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

Substitute $A B=A C$

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

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

$A C^{2}=A C^{2}+D C^{2}-2 A C \cdot D C+B D^{2}$

$2 A C \cdot D C=A C^{2}-A C^{2}+D C^{2}+B D^{2}$

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

Now, in $\triangle B D C$, we have

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

Therefore, $2 A C \cdot D C=D C^{2}+B D^{2}$

$2 A C . D C=B C^{2}$

Hence proved.


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