a) Vì \(\Delta ABC\) có \(AB=AC\Rightarrow\Delta ABC\)cân \(\Rightarrow\widehat{B}=\widehat{C}\)
Xét \(\Delta ABK\) và \(\Delta ACK\) có:
\(\begin{matrix}AB=AC\left(gt\right)\\\widehat{B}=\widehat{C}\left(cmt\right)\\BK=CK\left(gt\right)\end{matrix}\Rightarrow\Delta ABK=\Delta ACK\left(c-g-c\right)\)
b) Từ \(\Delta ABK=\Delta ACK\) \(\Rightarrow\widehat{BKA}=\widehat{CKA}\) ( 2 góc tương ứng )
Ta có: \(\widehat{BKA}+\widehat{CKA}=180^0\) ( 2 góc kề bù )
Mà \(\widehat{BKA}=\widehat{CKA}\)
\(\Rightarrow2\cdot\widehat{AKB}=180^0\)
\(\Rightarrow\widehat{AKB}=\dfrac{180^0}{2}=90^0\)
\(\Rightarrow BK\perp AK\)
Mà \(CK\perp BK\) \(\Rightarrow AK\) // \(EC\)
