Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {ADB} = \widehat {AEC} = {90^0}\\
AB = AC\\
\widehat Achung
\end{array} \right.\\
 \Rightarrow \Delta ADB = \Delta AEC\left( {ch - gn} \right)
\end{array}$

b) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BEC} = \widehat {CDB} = {90^0}\\
BCchung\\
\widehat {EBC} = \widehat {DCB}\left( {do:\widehat {ABC} = \widehat {ACB}} \right)
\end{array} \right.\\
 \Rightarrow \Delta EBC = \Delta DCB\left( {ch - gn} \right)\\
 \Rightarrow \widehat {ECB} = \widehat {DBC}\\
 \Rightarrow \widehat {OCB} = \widehat {OBC}\\
 \Rightarrow \Delta BOC \text{cân ở O}
\end{array}$

c) Ta có:

$\begin{array}{l}
\Delta ADB = \Delta AEC\left( {ch - gn} \right)\\
 \Rightarrow AD = AE\\
 \Rightarrow \Delta ADE\text{ cân ở A}\\
 \Rightarrow \widehat {ADE} = \dfrac{{{{180}^0} - \widehat {DAE}}}{2}\\
\text{Mà} \Delta ABC \text{cân ở A} \\
 \Rightarrow \widehat {ACB} = \dfrac{{{{180}^0} - \widehat {BAC}}}{2}\\
 \Rightarrow \widehat {ACB} = \widehat {ADE}\\
 \Rightarrow ED//BC
\end{array}$

d) Kẻ $CN//BE$ sao cho $N\in ME$

Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {BME} = \widehat {CMN}\left( {dd} \right)\\
BM = CM\\
\widehat {EBM} = \widehat {NCM}\left( {CN//BE} \right)
\end{array} \right.\\
 \Rightarrow \Delta BME = \Delta CMN\left( {g.c.g} \right)\\
 \Rightarrow \left\{ \begin{array}{l}
ME = MN\\
BE = CN
\end{array} \right.\\
 \Rightarrow \left\{ \begin{array}{l}
BE = CN\\
EN = 2EM\left( 1 \right)
\end{array} \right.
\end{array}$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
BE = NC\\
\widehat {BEC} = \widehat {NCE}\left( {do:BE//CN \Rightarrow CN \bot CE} \right)\\
CEchung
\end{array} \right.\\
 \Rightarrow \Delta BEC = \Delta NCE\left( {c.g.c} \right)\\
 \Rightarrow BC = NE\left( 2 \right)
\end{array}$

Từ $\left( 1 \right),\left( 2 \right) \Rightarrow 2EM = BC \Rightarrow EM = \dfrac{1}{2}BC$