A circle intersecting the lateral sides $A C$ and $C B$ of an isosceles triangle $A C B$ at points $P$ and $Q$, respectively, is circumscribed around triangle $A B Q$. Segments $A Q$ and $B P$ intersect at point $D$ such that $A Q: A D=4$ : 3. Find the area of triangle $D Q B$ if the area of triangle $P Q C$ is 3.
Consider two pairs of similar triangles $A C B$ and $P C Q, A D B$ and $Q D P$.

## Solution

Since

$$
\angle C P Q=180^{\circ}-\angle A P Q=\angle C B A=\angle C A B=180^{\circ}-\angle B Q P=\angle C Q P
$$

triangle $P C Q$ is also isosceles. Therefore, $P Q \| A B$. Triangle $A D B$ is similar to triangle $Q D P$ with a ratio of $\frac{A D}{D Q}=3$, so triangle $A C B$ is similar to triangle $P C Q$ with the same ratio of 3.

Thus,

$$
S_{\Delta \mathrm{ACB}}=9 \cdot S_{\Delta \mathrm{PCQ}}=27, S_{\mathrm{APQB}}=S_{\Delta \mathrm{ACB}}-S_{\Delta \mathrm{PCQ}}=27-3=24
$$

Since the heights of triangles $A Q B$ and $A P Q$, dropped to the bases $A B$ and $P Q$ respectively, are equal, the ratio of the areas of the triangles is equal to the ratio of the bases. Therefore,

$$
S_{\triangle \mathrm{AQB}}=\frac{3}{4} \cdot S_{\mathrm{APQB}}=\frac{3}{4} \cdot 24=18
$$

Thus,

$$
S_{\triangle \mathrm{DQB}}=\frac{D Q}{A Q} \cdot S_{\Delta \mathrm{AQB}}=\frac{1}{4} \cdot 18=\frac{9}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-14.jpg?height=509&width=443&top_left_y=757&top_left_x=813)

## Answer

$\frac{9}{2}$.