a,
Gọi $I$ là giao điểm $BC$ và $OM$
$\widehat{BAM}=\widehat{CAM}$
$\to \stackrel\frown{BM}=\stackrel\frown{CAM}$
$\to \widehat{BOM}=\widehat{COM}$
$\Delta OBM$ cân tại $O$ có: $\widehat{OMB}=\dfrac{180^o-\widehat{BOM}}{2}$
$\Delta MOC$ cân tại $O$ có: $\dfrac{OMC}=\dfrac{180^o-\widehat{COM}}{2}$
$\to \widehat{OMB}=\widehat{OMC}$
Mà $MB=MC$ (vì $\stackrel\frown{BM}=\stackrel\frown{MC}$), $MI$ chung
$\to\Delta BIM=\Delta CIM$ (c.g.c)
$\to BI=IC$
$\to OM\bot BC=I$
Mà $OM\bot DE\to BC//DE$
b,
$\widehat{EMC}=\widehat{MAC}=\dfrac{1}{2}sđ\stackrel\frown{MC}$
Mà $\widehat{BAM}=\widehat{CAM}$
$\to \widehat{MAB}=\widehat{EMC}$ $(1)$
$\widehat{CEM}=\dfrac{1}{2}(sđ\stackrel\frown{AM}-sđ\stackrel\frown{MC})= \dfrac{1}{2}(sđ\stackrel\frown{AM}-sđ\stackrel\frown{BM})=\dfrac{1}{2}sđ\stackrel\frown{AB}=\widehat{AMB}$ (2)
Từ $(1)(2)\to \Delta ABM\backsim\Delta MCE$ (g.g)
Tương tự ta có $\Delta AMC\backsim\Delta MDB$ (g.g)
