Chứng minh rằng: tam giác ADB=CEA

Ta có:

$\widehat{CAE}+\widehat{BAD}={90}^0$ (hai góc phụ nhau)

           $\widehat{ABD}+\widehat{BAD}={90}^0$ (hai góc phụ nhau)

           $\widehat{CAE}+\widehat{ACE}={90}^0$ (hai góc phụ nhau)

$\Rightarrow \widehat{ CAE}=\widehat{ABD}$  ; $\widehat{BAD}=\widehat{ACE}$

Xét $\Delta ADB$ và $\Delta CEA$có:

$\widehat{CAE}=\widehat{ABD}$; $AB=AC;$ $\widehat{BAD}=\widehat{ACE}$.

Vậy $\Delta ADB=\Delta CEA$ (g-c-g)