Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\left\{ \begin{array}{l}
AB = AC\\
\widehat {ABM} = \widehat {ACM}\\
MB = MC
\end{array} \right.\\
\Rightarrow \Delta AMB = \Delta AMC\left( {c.g.c} \right)
\end{array}$
b) Ta có:
$\begin{array}{l}
\Delta AMB = \Delta AMC\left( {c.g.c} \right)\\
\Rightarrow \widehat {MAB} = \widehat {MAC}
\end{array}$
$\to $ Tia $AM$ là tia phân giác góc $A$
c) Ta có:
$\begin{array}{l}
\Delta AMB = \Delta AMC\left( {c.g.c} \right)\\
\Rightarrow \widehat {AMB} = \widehat {AMC}\\
\Rightarrow \widehat {AMB} = \widehat {AMC} = \dfrac{{\widehat {AMB} + \widehat {AMC}}}{2} = \dfrac{{{{180}^0}}}{2} = {90^0}\\
\Rightarrow AM \bot BC
\end{array}$
d) Ta có:
$\begin{array}{l}
\Delta ABM;\widehat {AMB} = {90^0};AB = 5cm\\
\Rightarrow \left\{ \begin{array}{l}
BM = \dfrac{1}{2}BC = 3cm\\
AM = \sqrt {A{B^2} - A{M^2}} = 4cm
\end{array} \right.
\end{array}$
Vậy $BM=3cm;AM=4cm$
