Ta có \(\widehat{B_1}=\widehat{B_2}=\dfrac{1}{2}\widehat{ABC}\left(t/c.phân.giác\right);\widehat{C_1}=\dfrac{1}{2}\widehat{ABC}\left(gt\right)\)
Do đó \(\widehat{B_1}=\widehat{B_2}=\widehat{C_1}\)
Mà \(\widehat{B_1}+\widehat{B_3}=180^0\left(kề.bù\right);\widehat{C_1}+\widehat{C_2}=180^0\left(kề.bù\right)\)
Nên \(\widehat{B_3}=\widehat{C_2}\)
\(\left\{{}\begin{matrix}AB=CK\\EB=AC\\\widehat{B_3}=\widehat{C_2}\end{matrix}\right.\Rightarrow\Delta ABE=\Delta KCA\left(c.g.c\right)\Rightarrow AE=AK\)
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