Giải thích các bước giải:
a.Ta có $\Delta ABC$ cân tại $A$
$\to \widehat{ABC}=\widehat{ACB}$
Mà $\widehat{DME}=\widehat{ABC}\to \widehat{DME}=\widehat{ACB}=\widehat{MCE}$
$\to 180^o-\widehat{DMB}-\widehat{EMC}=180^o-\widehat{EMC}-\widehat{CEM}$
$\to \widehat{DMB}=\widehat{CEM}$
$\to \widehat{BDM}=180^o-\widehat{DBM}-\widehat{DMB}=180^o-\widehat{ECM}-\widehat{MEC}=\widehat{CME}$
b.Xét $\Delta DBM, \Delta CEM$ có:
$\widehat{BDM}=\widehat{CME}$
$\widehat{DBM}=\widehat{ECM}$
$\to \Delta BDM\sim\Delta CME(g.g)$
$\to$Tỉ số đồng dạng $k=\dfrac{BD}{CM}=\dfrac{DM}{ME}=\dfrac{BM}{CE}$