Ta có: $\widehat{DIC}=180^{\circ}-\widehat{AIC}=\widehat{IAC}+\widehat{ICA}=\frac{\widehat{A}+\widehat{C}}{2}$
Ta có $\widehat{BIH}=90^{\circ}-\frac{\widehat{B}}{2}=\frac{180^{\circ}-\widehat{B}}{2}=\frac{\widehat{A}+\widehat{C}}{2}=\widehat{DIC}$
Suy ra $\widehat{BIH}=\widehat{CID}$
