\(\widehat{A}+\widehat{B}+\widehat{C}=180^O\)
⇒\(\widehat{B}+\widehat{C}=180^0-\widehat{A}\)
⇒ \(\widehat{\dfrac{B}{2}}+\dfrac{\widehat{C}}{2}=90^0-\dfrac{\widehat{A}}{2}\)
⇔\(\widehat{BIC}=180^0\left(90^0-\dfrac{\widehat{A}}{2}\right)\)\(=90^0+\dfrac{\widehat{BIC}}{2}\left(đpcm\right)\)
