Chứng minh EC//AK biết đường thẳng qua C vuông góc với BC cắt AB tại E

a/ Xét \(\Delta AKB;\Delta AKC\) có :

\(\left\{{}\begin{matrix}AB=AC\\BK=KB\\AKchung\end{matrix}\right.\)

\(\Leftrightarrow\Delta AKB=\Delta AKC\left(c-c-c\right)\)

\(\Leftrightarrow\widehat{AKB}=\widehat{AKC}\)

Mà \(\widehat{AKB}+\widehat{AKC}=180^0\)

\(\Leftrightarrow\widehat{AKB}=\widehat{AKC}=\dfrac{180^0}{2}=90^0\)

\(\Leftrightarrow AK\perp BC\)

b/ Ta có :

\(\left\{{}\begin{matrix}AK\perp BC\\EC\perp BC\end{matrix}\right.\) \(\Leftrightarrow AK\backslash\backslash EC\)

c/ Ta có :

\(\widehat{BAC}+\widehat{CAE}=180^0\)

\(\Leftrightarrow\widehat{CAE}=180^0-\widehat{BAC}\)

\(\Leftrightarrow\widehat{CAE}=90^0\)

\(\Leftrightarrow\widehat{BAC}=\widehat{CAE}=90^0\)

Ta có \(\Delta ABC\) vuông cân tại A

\(\Leftrightarrow\widehat{CBA}=\widehat{ACB}=45^0\)

Xét \(\Delta CAE;\Delta ACB\) có :

\(\left\{{}\begin{matrix}ACchung\\\widehat{BAC}=\widehat{CAE}=90^0\\\widehat{CBA}=\stackrel\frown{ACB}\end{matrix}\right.\)

\(\Leftrightarrow\Delta CAE=\Delta ACB\left(ch-gn\right)\)

\(\Leftrightarrow CE=CB\)