Length of line segment in cyclic geometry

Since BC is the diameter of the circle: $\angle BPC = 90^\circ$ and $\angle BQC = 90^\circ$.

Thus, $CP \perp AB$ and $BQ \perp AC$. CP and BQ are altitudes of $\triangle ABC$.

The area of $\triangle ABC$ can be expressed in two ways: $\text{Area} = \frac{1}{2} \times AB \times CP = \frac{1}{2} \times AC \times BQ$ $\frac{1}{2} \times 30 \times 20 = \frac{1}{2} \times 25 \times BQ$ $600 = 25 \times BQ \implies BQ = 24\text{ cm}$