Question

In triangle $ABC$, $AB=AC$ and $BD$ is perpendicular to $AC$. Prove that
${BD}^2-{CD}^2=2CD\times AD$

Answer 1

In $△ABD,A{B}^2=A{D}^2+B{D}^2$

$⟹A{C}^2=A{D}^2+B{D}^2$

$(AD+CD{)}^2=A{D}^2+B{D}^2$

$A{D}^2+C{D}^2+2AD\times CD=A{D}^2+B{D}^2⟹B{D}^2-C{D}^2=2CD\times AD$