4. A right-angled triangle has legs with lengths $60 \mathrm{~cm}$ and $80 \mathrm{~cm}$. Let $M$ and $N$ be the midpoints of these legs. The circle with diameter $M N$ intersects the hypotenuse at points $P$ and $Q$. Determine the length of the segment $P Q$.
Solution. The given triangle has a hypotenuse $AB$ with a length of $100 \, \text{cm}$, so the midline $MN$ has a length of $50 \, \text{cm}$ (see the diagram). The given circle passes through the vertex $C$ of the right angle because the angle subtended by a diameter is a right angle. The point symmetric to $C$ with respect to the line $MN$ lies on the hypotenuse, so since the circle is symmetric with respect to the diameter $MN$, we conclude that this point lies on the circle. Therefore, it is one of the intersection points of the hypotenuse and the circle, say point $P$. Due to the symmetry, the segment $CP$ is perpendicular to $MN$, and thus to $AB$. Therefore, $\measuredangle CPQ = 90^\circ$. From this, it follows that the segment $CQ$ is a diameter of the circle, and its length is $50 \, \text{cm}$. For the area of triangle $ABC$, we have $P = \frac{60 \cdot 80}{2} = 2400$ and $P = \frac{AB \cdot CP}{2} = 50 \cdot CP$, so $50 \cdot CP = 2400$, i.e., $CP = 48 \, \text{cm}$.

Now, from the right triangle $CPQ$, we get

$$
PQ = \sqrt{CQ^2 - CP^2} = \sqrt{50^2 - 48^2} = \sqrt{196} = 14 \, \text{cm}
$$

which is what we needed to determine.