B4. A parallelogram $A B C D$ is cut by a line $m$. From the four vertices $A, B, C$ and $D$, we drop perpendiculars to line $m$. The feet of these perpendiculars are $P, Q, R$ and $S$ respectively. Point $S$ is also exactly the intersection of line $m$ and $A B$. The lengths of segments $A P, B Q$ and $D S$ are 6, 7, and 25, respectively.

What is the length of $C R$? Note: the diagram is not to scale.

![](https://cdn.mathpix.com/cropped/2024_04_17_8d46f403ab663a3b3700g-1.jpg?height=311&width=534&top_left_y=2323&top_left_x=1321)
B4. 12 We will prove that:

$$
|A P|+|B Q|+|C R|=|D S| .
$$

From this, the required length follows:

$$
|C R|=|D S|-|A P|-|B Q|=25-6-7=12 .
$$

Let $T$ be the foot of the perpendicular from $A$ to $D S$. It follows that $A P S T$ is a rectangle, so $|S T|=|A P|$.

![](https://cdn.mathpix.com/cropped/2024_04_17_351e9d9fc3725f0ab9f4g-2.jpg?height=389&width=670&top_left_y=1093&top_left_x=1178)

Now it remains to prove that $|D T|=|B Q|+|C R|$. Let $U$ be the foot of the perpendicular from $B$ to the line through $C$ and $R$. Since $B U R Q$ is a rectangle, it follows that $|R U|=|B Q|$, so $|C U|=|B Q|+|C R|$. Triangles $B U C$ and $A T D$ are similar, because the sides of the two triangles are pairwise parallel. The triangles are also congruent, because $|A D|=|B C|$. It follows that $|D T|=|C U|$ and thus that $|D T|=|B Q|+|C R|$.