Xét tam giác ABC có:
$\widehat{ABC}+\widehat{BAC}+\widehat{BCA}=180^{\circ}$
$\widehat{ABC}=180^{\circ}-\widehat{BAC}-\widehat{BCA}$ (1)
Xét tam giác ABD có:
$\widehat{ABD}+\widehat{BAD}+\widehat{BDA}=180^{\circ}$
$\widehat{ABD}=180^{\circ}-\widehat{BAD}-\widehat{BDA}$ (2)
Mà $\widehat{BAC}=\widehat{BAD};\widehat{BCA}=\widehat{BDA} $ (3)
Từ (1), (2), (3) ta suy ra $\widehat{ABC}=\widehat{ABD}$
Xét $\Delta ABC$ và $\Delta ABD$ có:
$\widehat{ABC}=\widehat{ABD}$ (chứng minh trên)
AB chung
$\widehat{BAC}=\widehat{BAD}$ (giả thiết)
Do đó, $\Delta ABC = \Delta ABD$ (g . c . g).