The triangles $A B C$ and $A B D$ are inscribed in the same semicircle with diameter $A B$, which measures $15 \mathrm{~cm}$. A perpendicular is drawn from $D$ to $A B$ intersecting $A B$ at $P$, the segment $A C$ at $Q$, and the extension of the side $B C$ at $R$. Additionally, $P R=\frac{40}{3} \mathrm{~cm}$ and $P Q=$ $\frac{15}{4} \mathrm{~cm}$.

a) Verify that $P Q \cdot P R=B P \cdot A P$.

b) Find the measure of the segment $D P$.
Solution

![](https://cdn.mathpix.com/cropped/2024_05_01_12f29789e10e6749ebb9g-11.jpg?height=608&width=722&top_left_y=1798&top_left_x=739)
a) Since $AB$ is a diameter, it follows that $\angle ACB = 90^{\circ}$. Therefore,

$$
\angle PRB = 90^{\circ} - \angle PBR = \angle BAC
$$

The triangles $APQ$ and $PRB$ are similar because they have the same angle measures. Therefore,

$$
\begin{aligned}
\frac{PQ}{AP} & = \frac{BP}{PR} \\
PQ \cdot PR & = BP \cdot AP
\end{aligned}
$$

b) The right triangles $ADP$ and $ABD$ are similar because they share one of the acute angles. The same applies to the right triangles $ABD$ and $DBP$. Therefore, $ADP$ is similar to $DBP$, and thus

$$
\begin{aligned}
\frac{DP}{AP} & = \frac{BP}{DP} \\
DP^2 & = AP \cdot BP
\end{aligned}
$$

From the previous item, it follows that

$$
\begin{aligned}
DP^2 & = PQ \cdot PR \\
DP & = \sqrt{\frac{40}{3} \cdot \frac{15}{4}} \\
& = 5 \sqrt{2}
\end{aligned}
$$