1) Xét $ΔABD$ và $ΔACB$ có:
$\widehat{A}:$ góc chung
$\widehat{ABD} =\widehat{ACB} \, (gt)$
Do đó $ΔABD\sim ΔACB \, (g.g)$
$\Rightarrow \dfrac{AB}{AC} = \dfrac{AD}{AB} = \dfrac{BD}{BC}$
2) Ta có: $\dfrac{AB}{AC} = \dfrac{AD}{AB}$ (câu a)
$\Rightarrow AB.AB = AD.AC$
hay $AB^2 - AD.AC$