In quadrilateral $A B C D$, side $A B$ is equal to side $B C$, diagonal $A C$ is equal to side $C D$, and $\angle A C B = \angle A C D$. The radii of the circles inscribed in triangles $A C B$ and $A C D$ are in the ratio $3: 4$. Find the ratio of the areas of these triangles.
Denote $\angle A C B=\angle A C D=2 \alpha$ and find $\cos 2 \alpha$, using the fact that the distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle.

## Solution

Let $\angle A C B=\angle A C D=2 \alpha, A C=C D=x, r$ and $R$ be the radii of the circles inscribed in triangles $A B C$ and $A C D$ respectively, $O$ and $Q$ be their centers, $M$ and $N$ be the midpoints of $A C$ and $A D$ respectively, $p_{1}$ and $p_{2}$ be the semiperimeters of triangles $A B C$ and $A D C$ respectively, and $K$ be the point of tangency of the inscribed circle of triangle $A D C$ with side $C$. Then

$$
C K=p_{2}-A D, C M=p_{1}-A B, C K=R \cot \alpha, C M=r \cot \alpha .
$$

Therefore,

$$
\frac{p_{2}-A D}{p_{1}-A B}=\frac{R}{r}=\frac{4}{3}
$$

or

$$
\frac{x-\frac{1}{2} A D}{\frac{x}{2}}=\frac{4}{3}
$$

From this equation, we find that $A D=\frac{2 x}{3}$. Therefore,

$$
\begin{aligned}
& \sin \alpha=\frac{D N}{C D}=\frac{1}{3}, \cos \alpha=\frac{2 \sqrt{2}}{3} \\
& \cos 2 \alpha=\frac{7}{9}, B C=\frac{C M}{\cos 2 \alpha}=\frac{9 x}{14}
\end{aligned}
$$

Then $\frac{p_{1}}{p_{2}}=\frac{6}{7}$. Therefore,

$$
\frac{S_{\triangle A B C}}{S_{\triangle A D C}}=\frac{r p_{1}}{R p_{2}}=\frac{3}{4} \cdot \frac{p_{1}}{p_{2}}=\frac{9}{14}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-12.jpg?height=483&width=688&top_left_y=2194&top_left_x=685)

## Answer

The points of tangency of the inscribed circle of a triangle are connected by segments, and heights are drawn in the resulting triangle. Prove that the lines connecting the bases of these heights are parallel to the sides of the original triangle.

## Hint

Apply the theorem about the angle between a tangent and a chord.

## Solution

Let $M, N$ and $K$ be the points of tangency of the circle inscribed in triangle $A B C$ with sides $B C, A C$ and $A B$ respectively; $M P$ and $K Q$ be the heights of triangle $M N K$. From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle M N C=\angle M K N=\angle P Q N
$$

Therefore, $P Q \| A C$.

Let $M, N$ and $K$ be the points of tangency of the circle inscribed in triangle $A B C$ with sides $B C, A C$ and $A B$ respectively; $M P$ and $K Q$ be the heights of triangle $M N K$. From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle M N C=\angle M K N=\angle P Q N
$$

Therefore, $P Q \| A C$.

Let $M, N$ and $K$ be the points of tangency of the circle inscribed in triangle $A B C$ with sides $B C, A C$ and $A B$ respectively; $M P$ and $K Q$ be the heights of triangle $M N K$. From the theorem about the angle between a tangent and a chord, it follows that

$$
\angle M N C=\angle M K N=\angle P Q N
$$

Therefore, $P Q \| A C$.

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-13.jpg?height=437&width=794&top_left_y=1465&top_left_x=632)

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