Ta có: ΔABC cân tại A
=>\(\widehat{ACB}=\dfrac{180^0-\widehat{BAC}}{2}\)
Ta có: \(\widehat{xCA}+\widehat{ACB}=\widehat{xCB}=90^0\)
=>\(\widehat{xCA}=90^0-\widehat{ACB}\)
=>\(\widehat{xCA}=90^0-\dfrac{180^0-\widehat{BAC}}{2}\)
=>\(\widehat{xCA}=90^0-90^0+\dfrac{1}{2}\cdot\widehat{BAC}=\dfrac{1}{2}\cdot\widehat{BAC}\)