$\text{a) Gọi:}$ $MF\perp BD, MG\perp DE, MH\perp CE$

$\text{Ta có:}$ $\Delta ABC$ $\text{cân tại}$ $A, M$ $\text{là trung điểm BC}$

$\to AM$ $\text{là phân giác}$ $\widehat{BAC}$

$\text{Mà}$ $MF\perp AB, MH\perp AC\to MF=MH$

$\text{Lại có:}$  $DM$ $\text{là phân giác}$  $\widehat{BDE}, MF\perp BD, MG\perp DE$

$\to MF=MG$
$\to MF=MG=MH$

$Do$ $MG\perp ED, MH\perp EC$

$\to EM$ $\text{là phân giác}$ $\widehat{DEC}$

$\text{b) Ta có:}$ $\Delta ABC$ $cân tại$ $A\to \widehat{ABC}=\widehat{ACB}$

$\to \widehat{DBM}=\widehat{ECM}$

$\text{Lại có:}$ 

$\widehat{DME}=\widehat{DMG}+\widehat{GME}=\dfrac12\widehat{GMF}+\dfrac12\widehat{GMH}=\dfrac12\widehat{FMH}=\widehat{FMA}=90^o-\widehat{FMB}=\widehat{FBM}=\widehat{DBM}$

$\text{Mà}$ $\widehat{BDM}=\widehat{MDE}$

$\to\Delta DBM\sim\Delta DME(g.g)$

$\to \widehat{DMB}=\widehat{DEM}=\widehat{MEC}$

$\text{Kết hợp}$

$\widehat{DBM}=\widehat{ECM}$

$\to \Delta BDM\sim\Delta CME(g.g)$

$\text{c) Từ câu (b)}$ 

$\to\dfrac{BD}{CM}=\dfrac{BM}{CE}$

$\to BD\cdot CE=MB\cdot MC=\dfrac12BC\cdot \dfrac12CB=a^2$