Giải thích các bước giải:
Kẻ $AD//BC, D\in BM$
Vì $M$ là trung điểm $AC$
$\to \dfrac{AD}{BC}=\dfrac{MD}{MB}=\dfrac{MA}{MC}=1\to AD=BC, MB=MD$
Ta có: $\dfrac{IB}{ID}=\dfrac{BH}{AD}=\dfrac{AC}{BC}$
$\to \dfrac{IB}{IB+ID}=\dfrac{AC}{AC+BC}$
$\to \dfrac{IB}{BD}=\dfrac{AC}{AC+BC}$
$\to \dfrac{IB}{2MB}=\dfrac{AC}{AC+BC}$
$\to \dfrac{IB}{MB}=\dfrac{2AC}{AC+BC}$
$\to \dfrac{IB}{MB-IB}=\dfrac{2AC}{AC+BC-2AC}$
$\to \dfrac{IB}{IM}=\dfrac{2AC}{BC-AC}$
$\to \dfrac{IB}{IM}=\dfrac{2AC}{BC-BH}$
$\to \dfrac{IB}{IM}=\dfrac{2AC}{CH}$
$\to \dfrac{IB}{IM}=\dfrac{2AC.CB}{CH.CB}$
Mà $\widehat{CHA}=\widehat{CAB}=90^o,\widehat{ACH}=\widehat{ACB}$
$\to\Delta CHA\sim\Delta CAB(g.g)$
$\to \dfrac{CA}{CB}=\dfrac{CH}{CA}\to CH.CB=CA^2$
$\to \dfrac{IB}{IM}=\dfrac{2AC.CB}{CA^2}$
$\to \dfrac{IB}{IM}=\dfrac{CB}{\dfrac{CA}2}$
$\to \dfrac{IB}{IM}=\dfrac{CB}{CM}$
$\to CI$ là phân giác $\widehat{MCB}$
$\to CI$ là phân giác $\widehat{ACB}$
