Giải thích các bước giải:

 a) Ta có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {PMQ} = \widehat {QRS} = {90^0}\\
\widehat {PQM} = \widehat {QSR}\left( {PQ//SR} \right)
\end{array} \right.\\
 \Rightarrow \Delta PMQ \sim \Delta QRS\left( {g.g} \right)
\end{array}$

b) Ta có:

$\begin{array}{l}
\Delta PQS;\widehat P = {90^0};PQ = 16cm;PS = 12cm\\
 \Rightarrow \left\{ \begin{array}{l}
QS = \sqrt {P{Q^2} + P{S^2}}  = 20cm\\
PM = \dfrac{{2{S_{PQS}}}}{{QS}} = \dfrac{{PQ.PS}}{{QS}} = 9,6cm
\end{array} \right.
\end{array}$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat Qchung\\
\widehat {PMQ} = \widehat {SPQ} = {90^0}
\end{array} \right.\\
 \Rightarrow \Delta PMQ \sim \Delta SPQ\left( {g.g} \right)\\
 \Rightarrow \dfrac{{PM}}{{SP}} = \dfrac{{MQ}}{{PQ}}\\
 \Rightarrow MQ = \dfrac{{PQ.PM}}{{SP}} = \dfrac{{16.9,6}}{{12}} = 12,8cm\\
 \Rightarrow MS = PS - MQ = 7,2cm
\end{array}$

Vậy $QS = 20cm;PM = 9,6cm;MQ = 12,8cm;MS = 7,2cm$

c) Ta có;

$\begin{array}{l}
\Delta PMQ \sim \Delta SPQ\left( {g.g} \right)\\
 \Rightarrow \dfrac{{MQ}}{{PQ}} = \dfrac{{PQ}}{{SQ}}\\
 \Rightarrow P{Q^2} = MQ.SQ
\end{array}$

Lại có:

$\begin{array}{l}
\left\{ \begin{array}{l}
\widehat {PMS} = \widehat {QPS} = {90^0}\\
\widehat Schung
\end{array} \right.\\
 \Rightarrow \Delta PMS \sim \Delta QPS\left( {g.g} \right)\\
 \Rightarrow \dfrac{{PS}}{{QS}} = \dfrac{{MS}}{{PS}}\\
 \Rightarrow P{S^2} = QS.MS
\end{array}$

d) Ta có:

$\begin{array}{l}
\Delta PMS \sim \Delta QPS\\
\Delta PMQ \sim \Delta SPQ\\
 \Rightarrow \Delta PMS \sim \Delta QMP\\
 \Rightarrow \dfrac{{{S_{PMS}}}}{{{S_{QMP}}}} = {\left( {\dfrac{{PM}}{{MQ}}} \right)^2} = {\left( {\dfrac{{9,6}}{{12,8}}} \right)^2} = \dfrac{9}{{16}}
\end{array}$

Lại có:

$\begin{array}{l}
\Delta PMQ \sim \Delta QRS\left( {g.g} \right)\\
 \Rightarrow \dfrac{{{S_{PMQ}}}}{{{S_{QRS}}}} = {\left( {\dfrac{{PM}}{{QR}}} \right)^2} = {\left( {\dfrac{{9,6}}{{12}}} \right)^2} = \dfrac{{16}}{{25}}
\end{array}$