Question
In $\triangle ABC, AB = AC$ and $BD$ is perpendicular to $AC.$Prove that: $BD^2- CD^2= 2CD \times AD.$
✓
Answer
In right-angled $\triangle ADB$,
$A B^2=A D^2+B D^2$
$\Rightarrow A D^2=A B^2-B D^2 .....(i)$
$ A C=A D+D C$
$ \Rightarrow A C^2=(A D+D C)^2$
$ \Rightarrow A C^2=A D^2+D C^2+2 A D \times D C$
$ \Rightarrow AC ^2= AB ^2- BD ^2+ DC ^2+2 AD \times DC \ldots[$ From $(i)]$
$ \Rightarrow A C^2=A C^2-B D^2+D C^2+2 A D \times D C \ldots[A B=A C]$
$ \Rightarrow BD ^2- DC ^2=2 AD \times DC.$
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