Giải thích các bước giải:
Gọi $AB\cap IE=F$
Xét $\Delta MAB,\Delta MCD$ có:
$MA=MD$
$\widehat{AMB}=\widehat{CMD}$
$MB=MC$
$\to\Delta AMB=\Delta DMC(c.g.c)$
$\to \widehat{MAB}=\widehat{MDC}\to AB//CD$
$\to CD\perp AC$ vì $AB\perp AC$
Xét $\Delta AFC, \Delta FCI$ có:
$\widehat{AFC}=\widehat{FCI}$ vì $AB//CD$
Chung $CF$
$\widehat{ACF}=\widehat{CFI}$ vì $EI//AC$
$\to\Delta ACF=\Delta IFC(g.c.g)$
$\to AF=CI$
Mà $CI=CA\to AF=AC$
Xét $\Delta AEF,\Delta ABC$ có:
$\widehat{EFA}=\widehat{BAC}(=90^o)$
$AF=AC$
$\widehat{EAF}=\widehat{BAH}=90^o-\widehat{HAC}=\widehat{ACH}=\widehat{ACB}$
$\to\Delta AEF=\Delta CBA(g.c.g)$
$\to AE=BC$