<s>[INST] What is the value of the PM segment in the figure below? For reference:
Given a rectangle, ABCD (AB > BC) by B draws a perpendicular to AC which
intercepts "P" to CD, and an "M" the perpendicular to BD drawn by B.
$\space$ The prolongation of DM and BC intersect at Q.
Calculate PM if DQ =$ 17$ and BP = $9$. (answer: $4$)

My progresss:
$$\triangle BCP \sim \triangle QCD\\
\frac{BC}{CQ}=\frac{PC}{2PC}=\frac{PB}{DQ}\rightarrow\\
\frac{BC}{17}=\frac{1}{2}=\frac{9}{DQ} \therefore BC = \frac{17}{2}; DQ=18\\
\triangle DCQ: DC^2+CQ^2 = DQ^2 \rightarrow\\
DC^2+17^2 =18^2 \therefore DC =\sqrt35\\
\text{draw} ~MI \perp BQ \implies \triangle MBQ (\text{isosceles})\\
\triangle BMI \sim \triangle DCQ (AA):\\
BI = \frac{BQ}{2} = \frac{\frac{17}{2}+17}{2} =\frac{51}{4}\\\\
\frac{BI}{17}=\frac{BP+PM}{DQ}\rightarrow \frac{51}{68}=\frac{9+PM}{18} \therefore \boxed{ PM=4,5}$$

I didn't see where my mistake is..(book answer = 4)
 [/INST] 
Hints: If you show that In figure M is midpoint of QT and DM=PM, then we have:
In right angled triangle TBQ, BM=MQ so:
$17-PM=9+PM\Rightarrow PM=4$
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