Question

Answer 1

Ta có: \(\widehat{ABC}=2\widehat{ABD}\) (BD là phân giác \(\widehat{ABC}\)).

           \(\widehat{ABC}=2\widehat{ACB}\left(gt\right).\)

\(\Rightarrow\widehat{ABD}=\widehat{ACB}.\)

Mà \(\left\{{}\begin{matrix}\widehat{ABE}=180^o-\widehat{ABD}.\\\widehat{KCA}=180^o-\widehat{ACB}.\end{matrix}\right.\)

\(\Rightarrow\widehat{ABE}=\widehat{KCA}.\)

Xét \(\Delta ABE\) và \(\Delta KCA:\)

\(AB=CK\left(gt\right).\\ BE=AC\left(gt\right).\\ \widehat{ABE}=\widehat{KCA}\left(cmt\right).\\ \Rightarrow\Delta ABE=\Delta KCA\left(c-g-c\right).\)

\(\Rightarrow AE=KA\) (2 cạnh tương ứng).