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).