[ Rhombi. Properties and characteristics ] [Area of a triangle (using two sides and the angle between them).]

In an acute-angled triangle $A B C$, point $D$ is chosen on side $A B$ such that $\angle D C A=45^{\circ}$. Point $D_{1}$ is symmetric to point $D$ with respect to line $B C$, and point $D_{2}$ is symmetric to point $D_{1}$ with respect to line $A C$ and lies on the extension of segment $B C$ beyond point $C$. Find the area of triangle $A B C$, if $B C=\sqrt{3} C D_{2}, A B=4$.
$\angle B C D=30^{\circ}$, quadrilateral $D B D_{1} C-$ is a rhombus.

## Solution

Let $Q$ and $P$ be the points of intersection of segments $D D_{1}$ and $D_{1} D_{2}$ with lines $B C$ and $A C$. Denote $\angle A C B=\gamma$. Then

$$
\begin{gathered}
\angle D_{1} C P=\angle D_{2} C P=\angle A C B=\gamma \\
\angle Q C D_{1}=\angle Q C D=\angle A C B-\angle A C D=\gamma-45^{\circ}
\end{gathered}
$$

Since

$$
\angle Q C D_{1}+\angle D_{1} C P+\angle D_{2} C P=180^{\circ}
$$

From this, we find that $\gamma=75^{\circ}$, and $\angle D C B=30^{\circ}$.

Let $C D_{2}=x$. Then

$$
\begin{gathered}
D C=D_{1} C=D_{2} C=x, B C=x \sqrt{3} \\
Q C=D C \cos 30^{\circ}=\frac{x \sqrt{3}}{2}, B Q=B C-Q C=\frac{x \sqrt{3}}{2}=Q C
\end{gathered}
$$

Therefore, $D B D_{1} C-$ is a rhombus. Then

$$
\begin{gathered}
\angle A B C=\angle D C B=30^{\circ} \\
\angle B A C=180^{\circ}-\angle A B C-\angle A C B=180^{\circ}-30^{\circ}-75^{\circ}=75^{\circ} .
\end{gathered}
$$

Therefore, triangle $A B C-$ is isosceles, $B C=A B=4$. Consequently,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \sin 30^{\circ}=\frac{1}{2} \cdot 4 \cdot 4 \cdot \frac{1}{2}=4
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_9466e5d28b3947d5c4a5g-32.jpg?height=411&width=920&top_left_y=1018&top_left_x=569)

## Answer

4. 

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