Giải thích các bước giải:

a.Vì $\Delta AEH$ vuông tại $E, I$ là trung điểm $AH$

$\to IE=IA=IH=\dfrac12AH$

Tương tự $IF=IA=IH=\dfrac12AH$

$\to IE=IF$

Tương tự $MA=MF$

$\to MI$ là trung trực $EF$

$\to IM\perp EF$

b.Từ a $\to IA=IH=IE=IF$

$\to \Delta IEF$ cân tại $I$

$\to \widehat{IFK}=\dfrac12(\widehat{IFE}+\widehat{IEF})$

$\to \widehat{IFK}=\dfrac12(\widehat{AFE}-\widehat{IFA}+\widehat{AEF}-\widehat{IEA})$

$\to \widehat{IFK}=\dfrac12(\widehat{AFE}+\widehat{AEF}-\widehat{IEA}-\widehat{IFA})$

$\to \widehat{IFK}=\dfrac12(\widehat{AFE}+\widehat{AEF}-\widehat{IAE}-\widehat{IAF})$

$\to \widehat{IFK}=\dfrac12(\widehat{AFE}+\widehat{AEF}-\widehat{FAE})$

$\to \widehat{IFK}=\dfrac12(180^o-2\widehat{FAE})$

$\to \widehat{IFK}=\dfrac12(180^o-2\widehat{BAC})$

$\to \widehat{IFK}=90^o-\widehat{BAE}=\widehat{ABE}=\widehat{FBH}$

Xét $\Delta AFH,\Delta ADB$ có:

Chung $\hat A$

$\hat F=\hat D(=90^o)$

$\to \Delta AFH\sim\Delta ADB(g.g)$

$\to \dfrac{AF}{AD}=\dfrac{AH}{AB}$

$\to \Delta AFD\sim\Delta AHB(c.g.c)$

$\to \widehat{ADF}=\widehat{ABH}$

$\to \widehat{IFK}=\widehat{IDF}$

$\to \Delta IKF\sim\Delta IFD(g.g)$

$\to \dfrac{IK}{IF}=\dfrac{IF}{ID}$

$\to IF^2=IK.ID$