a: Ta có: \(\widehat{OAD}=\dfrac{\widehat{BAD}}{2}\)
\(\widehat{ODA}=\dfrac{\widehat{ADC}}{2}\)
Do đó: \(\widehat{OAD}+\widehat{ODA}=\dfrac{1}{2}\left(\widehat{BAD}+\widehat{ADC}\right)\)
hay \(\widehat{OAD}+\widehat{ODA}=90^0\)
Xét ΔOAD có \(\widehat{OAD}+\widehat{ODA}=90^0\)
nên ΔOAD vuông tại O
hay AE\(\perp\)DB tại O