In triangle $ABC$, the lines containing the altitudes $AP$, $CR$, and $BQ$ (points $P$, $R$, and $Q$ lie on the lines containing the corresponding sides of triangle $ABC$) intersect at point $O$. Find the areas of triangles $ABC$ and $POR$, given that $RP$ is parallel to $AC$, $AC=4$, and $\sin \angle ABC=\frac{24}{25}$.
Prove that the given triangle is isosceles. Consider two cases: $\angle A B C = 90^{\circ}$.

## Solution

Since from points $P$ and $R$ the segment $A C$ is seen at a right angle, these points lie on a circle with diameter $AC$. Therefore, the trapezoid ACPR (Fig.1) (or ACRP (Fig.2)) is isosceles. Hence, triangle $ABC$ is also isosceles. Its equal angles at the base $AC$ are acute, and $BQ$ is both a median and an angle bisector.

Let $\angle A B C = \beta$. Suppose $\beta = 90^{\circ}$ (Fig. 2). Then

$$
\begin{gathered}
\cos \beta = -\frac{7}{25}, \operatorname{ctg} \frac{\beta}{2} = \frac{1 + \cos \beta}{\sin \beta} = \frac{1 - \frac{7}{25}}{\frac{24}{25}} = \frac{18}{24} = \frac{3}{4} \\
\operatorname{tg} \frac{\beta}{2} = \frac{4}{3}, \cos \frac{\beta}{2} = \frac{1}{\sqrt{1 + \left(\frac{4}{3}\right)^{2}}} = \frac{3}{5}.
\end{gathered}
$$

From the right triangle $AQB$, we find that

$$
BQ = AQ \cdot \operatorname{ctg} \angle ABQ = 2 \cdot \operatorname{ctg} \frac{\beta}{2} = 2 \cdot \frac{3}{4} = \frac{3}{2}
$$

Therefore,

$$
S_{\triangle ABC} = \frac{1}{2} \cdot AC \cdot BQ = \frac{1}{2} \cdot 4 \cdot \frac{3}{2} = 3
$$

From the right triangles OQC and ORC, we find that

$$
OC = \frac{CQ}{\cos \angle OCQ} = \frac{2}{\cos \frac{\beta}{2}} = \frac{2}{\frac{3}{5}} = \frac{10}{3},
$$

$$
PC = OC \cdot \sin \angle COP = OC \cdot \sin \beta = \frac{10}{3} \cdot \frac{24}{25} = \frac{16}{5}
$$

$$
OP = OC \cdot \cos \angle COP = OC \cdot \cos (180^{\circ} - \beta) = -OC \cdot \cos \beta = -\frac{10}{3} \cdot \left(-\frac{7}{25}\right) = \frac{14}{15}
$$

Therefore,

$$
S_{\triangle POC} = \frac{1}{2} \cdot PC \cdot OP = \frac{1}{2} \cdot \frac{16}{5} \cdot \frac{14}{15} = \frac{112}{75}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8027e864227276d12944g-44.jpg?height=1329&width=608&top_left_y=393&top_left_x=725)

## Answer

$\frac{16}{3}$ and $\frac{21}{25}$ or 3 and $\frac{112}{75}$.