$\text{a) Gọi:}$ $MF\perp BD, MG\perp DE, MH\perp CE$
$\text{Ta có:}$ $\Delta ABC$ $\text{cân tại}$ $A, M$ $\text{là trung điểm BC}$
$\to AM$ $\text{là phân giác}$ $\widehat{BAC}$
$\text{Mà}$ $MF\perp AB, MH\perp AC\to MF=MH$
$\text{Lại có:}$ $DM$ $\text{là phân giác}$ $\widehat{BDE}, MF\perp BD, MG\perp DE$
$\to MF=MG$
$\to MF=MG=MH$
$Do$ $MG\perp ED, MH\perp EC$
$\to EM$ $\text{là phân giác}$ $\widehat{DEC}$
$\text{b) Ta có:}$ $\Delta ABC$ $cân tại$ $A\to \widehat{ABC}=\widehat{ACB}$
$\to \widehat{DBM}=\widehat{ECM}$
$\text{Lại có:}$
$\widehat{DME}=\widehat{DMG}+\widehat{GME}=\dfrac12\widehat{GMF}+\dfrac12\widehat{GMH}=\dfrac12\widehat{FMH}=\widehat{FMA}=90^o-\widehat{FMB}=\widehat{FBM}=\widehat{DBM}$
$\text{Mà}$ $\widehat{BDM}=\widehat{MDE}$
$\to\Delta DBM\sim\Delta DME(g.g)$
$\to \widehat{DMB}=\widehat{DEM}=\widehat{MEC}$
$\text{Kết hợp}$
$\widehat{DBM}=\widehat{ECM}$
$\to \Delta BDM\sim\Delta CME(g.g)$
$\text{c) Từ câu (b)}$
$\to\dfrac{BD}{CM}=\dfrac{BM}{CE}$
$\to BD\cdot CE=MB\cdot MC=\dfrac12BC\cdot \dfrac12CB=a^2$
