Giải thích các bước giải:

a.Xét $\Delta AEC, \Delta AFB$ có:

Chung $\hat A$

$\widehat{AEC}=\widehat{AFB}(=90^o)$

$\to \Delta AEC\sim\Delta AFB(g.g)$

b.Vì $BF\perp AC, CE\perp AB, BF\cap CE=H$

$\to H$ là trực tâm $\Delta ABC$

$\to AH\perp BC=D$

Xét $\Delta AEH,\Delta ADB$ có:

Chung $\hat A$

$\widehat{AEH}=\widehat{ADB}(=90^o)$

 $\to \Delta AEH\sim\Delta ADB(g.g)$

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

$\to AE.AB=AH.AD$

c.Ta có: $DM//HE(\perp AB), DN//HF(\perp AC)$

$\to \dfrac{AE}{AM}=\dfrac{AH}{AD}=\dfrac{AF}{AN}$

$\to EF//MN$

$\to \widehat{AEF}=\widehat{AMN}\to \widehat{AEI}=\widehat{AMJ}$

Mà $\dfrac{AE}{AM}=\dfrac{EF}{MN}=\dfrac{2EI}{2MJ}=\dfrac{EI}{MJ}$

$\to \Delta AEI\sim\Delta AMJ(c.g.c)$

$\to \widehat{EAI}=\widehat{MAJ}$

$\to A, I, J$ thẳng hàng