Cho tam giác ABC (góc A = 90 độ), AB = AC Kẻ CE vuông góc AB tại E

* Hướng dẫn giải

a. $\left.\begin{array}{l}\widehat{A}+\widehat{ABD}={90}^0\\ \widehat{A}+\widehat{ACE}={90}^0\end{array}\right\}\Rightarrow \widehat{ABD}=\widehat{ACE}$

Xét $\Delta ABD$ và $\Delta ACE$ có

$\left.\begin{array}{l}\widehat{ABD}=\widehat{ACE}\\ AB=AC\\ \widehat{A}chung\end{array}\right\}\Rightarrow \Delta ABD=\Delta ACE\left(gcg\right)$

b. $\Delta ABD=\Delta ACE$ (cmt) => AE = AD

$\left.\begin{array}{l}AE+EB=AB\\ AD+DC=AC\\ AB=AC;AE=AD\end{array}\right\}\Rightarrow EB=DC$

Xét $\Delta BEO$ và $\Delta CDO$ có

$\left.\begin{array}{l}\widehat{ABD}=\widehat{ACE}\\ BE=DC\\ \widehat{BEO}=\widehat{CDO}\end{array}\right\}\Rightarrow \Delta BEO=\Delta CDO\left(gcg\right)$

c. $\Delta BEO=\Delta CDO\Rightarrow OB=OC$

Xét $\Delta AOB$ và $\Delta AOC$ có

$\begin{array}{l}\left.\begin{array}{l}OAchung\\ AB=AC\\ OB=OC\end{array}\right\}\Rightarrow \Delta AOB=\Delta AOC\left(gcg\right)\Rightarrow \widehat{BAO}=\widehat{CAO}\\ \end{array}$