Xét $∆ABD$ và $∆ACB$ có:
$\widehat{A}:$ góc chung
$\widehat{ABD}=\widehat{ACB}$ (cùng chắn $\mathop{BD}\limits^{\displaystyle\frown}$)
Do đó $∆ABD\sim ∆ACB\, (g.g)$
$\to \dfrac{AB}{AC}=\dfrac{BD}{BC}$
$\to \dfrac{AB^2}{AC^2}=\dfrac{BD^2}{BC^2}$
$\to \dfrac{AC.AD}{AC^2}=\dfrac{BD^2}{BC^2}$
$\to \dfrac{AD}{AC}=\dfrac{BD^2}{BC^2}$