Giải thích các bước giải:

a.Xét $\Delta BEM,\Delta CFN$ có:

$\hat M=\hat N(=90^o)$

$BM=CN$

$\hat B=\widehat{ACB}=\widehat{FNC}$

$\to \Delta MBE=\Delta NCF(g.c.g)$

$\to ME=NF$

b.Xét $\Delta EMB,\Delta EMD$ có:

Chung $EM$

$\widehat{EMD}=\widehat{EMB}(=90^o)$

$\widehat{EDM}=\widehat{ACB}=\widehat{ABC}=\hat B$

$\to \Delta EMD=\Delta EMB(g.c.g)$

$\to MD=MB$

c.Từ b $\to ED=EB$

         a $\to EB=CF$

$\to ED=CF$

Xét $\Delta ODE,\Delta OCF$ có:

$\widehat{OED}=\widehat{OFC}$ vì $DE//CF$

$DE=CF$

$\widehat{EOD}=\widehat{COF}$

$\to \Delta ODE=\Delta OCF(g.c.g)$

$\to OE=OF$