Giải thích các bước giải:
a) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 12cm;AC = 16cm\\
\Rightarrow \left\{ \begin{array}{l}
BC = \sqrt {A{B^2} + A{C^2}} = \sqrt {{{12}^2} + {{16}^2}} = 20cm\\
\tan B = \dfrac{{AC}}{{AB}} = \dfrac{4}{3} \Rightarrow \widehat B = 53,{13^0}\\
\widehat C = {90^0} - \widehat B = 36,{87^0}
\end{array} \right.
\end{array}$
Vậy ${BC = 20cm;\widehat B = 53,{{13}^0};\widehat C = 36,{{87}^0}}$
b) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AB = 12cm;AC = 16cm;BC = 20cm\\
\Rightarrow AM.BC = AB.AC\\
\Rightarrow AM = \dfrac{{AB.AC}}{{BC}} = \dfrac{{12.16}}{{20}} = 9,6cm
\end{array}$
Lại có:
$\begin{array}{l}
\Delta ABM;\widehat {AMB} = {90^0};AB = 12cm;AM = 9,6cm\\
\Rightarrow BM = \sqrt {A{B^2} - A{M^2}} = \sqrt {{{12}^2} - 9,{6^2}} = 7,2cm
\end{array}$
Vậy $AM = 9,6cm;BM = 7,2cm$
c) Ta có:
$\begin{array}{l}
\Delta ABM;\widehat {AMB} = {90^0};ME \bot AB = E\\
\Rightarrow A{M^2} = AE.AB\left( 1 \right)\\
\Delta ACM;\widehat {AMC} = {90^0}\\
\Rightarrow A{M^2} = A{C^2} - M{C^2}\left( 2 \right)
\end{array}$
Từ $\left( 1 \right),\left( 2 \right) \Rightarrow AE.AB = A{C^2} - M{C^2}$ (ĐPCM)
d) Ta có:
$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AM \bot BC = M\\
\Rightarrow A{M^2} = MB.MC\left( 3 \right)
\end{array}$
Lại có:
Xét $\Delta AEM;\Delta CMA$ có:
$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {AEM} = \widehat {CMA} = {90^0}\\
\widehat {EAM} = \widehat {MCA}\left( { + \widehat {MAC} = {{90}^0}} \right)
\end{array} \right.\\
\Rightarrow \Delta AEM \sim \Delta CMA\left( {g.g} \right)\\
\Rightarrow \dfrac{{AM}}{{CA}} = \dfrac{{EM}}{{MA}}\\
\Rightarrow A{M^2} = EM.AC\left( 4 \right)
\end{array}$
Từ $\left( 1 \right),\left( 3 \right),\left( 4 \right) \Rightarrow AE.AB = MB.MC = EM.AC$ (ĐPCM)
