Giải thích các bước giải:

a.Xét $\Delta MAB,\Delta MCD$ có:

$MA=MD$

$\widehat{AMB}=\widehat{CMD}$

$MB=MC$

$\to \Delta AMB=\Delta DMC(c.g.c)$

$\to \widehat{MAB}=\widehat{MDC}$

$\to AB//CD$

Do $AB\perp CD\to CD\perp AC$

b.Xét $\Delta FAI,\Delta CAI$ có:

$\widehat{FIA}=\widehat{IAC}$ vì $FI//AC$

Chung $AI$

$\widehat{FAI}=\widehat{AIC}$ vì $AF//CI$

$\to \Delta FAI=\Delta CIA(g.c.g)$

$\to FA=CI, FI=AC$

Xét $\Delta GAF, \Delta GCI$ có:

$GF=GC$ vì $G$ là trung điểm $CF$

$\widehat{GFA}=\widehat{GCI}$ vì $FA//CI$

$AF=CI$

$\to \Delta GAF=\Delta GIC(c.g.c)$

$\to \widehat{FGA}=\widehat{IGC}$

$\to A, G, I$ thẳng hàng

c.Ta có: $AF=CI, CA=FI, CA=CI$

$\to AF=FI=IC=CA$

Xét $\Delta AFE,\Delta ABC$ có:

$\widehat{EFA}=\widehat{ACD}(=90^o)$

$AF=AC$

$\widehat{FAE}=\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}=\widehat{ACB}$

$\to \Delta FEA=\Delta ABC(g.c.g)$

$\to AE=BC$