In parallelogram $A B C D$, angle $A$ is obtuse, $A D>A B, A D=7$. Point $A_{1}$ is symmetric to point $A$ with respect to line $B D$, and point $A_{2}$ is symmetric to point $A_{1}$ with respect to line $A C$ and lies on diagonal $B D$. Find the area of parallelogram $A B C D$, if $B A_{2}=\frac{4}{5} B D$.
Prove that triangle $A A_{1} A_{2}$ is equilateral and apply the Law of Cosines.

## Solution

From the properties of axial symmetry, it follows that $A_{2} A = A_{2} A_{1}$ and $A A_{2} = A A_{1}$. Therefore, triangle $A A_{1} A_{2}$ is equilateral, and point $O$, the intersection of the diagonals of parallelogram $A B C D$, is the center of this triangle.

Let $B D = 5 x$. Then

$$
A_{2} D = x, A O = O A_{2} = O D - A_{2} D = \frac{5 x}{2} - x = \frac{3 x}{2}, \angle A O D = 120^{\circ}
$$

By the Law of Cosines,

$$
A D^{2} = O A^{2} + O D^{2} - 2 O A \cdot O D \cos 120^{\circ}, \text{ or } 49 = \frac{9 x^{2}}{4} + \frac{25 x^{2}}{4} + \frac{15 x^{2}}{4}.
$$

From this, we find that $x = 2$. Then $B D = 10, A C = 2 A O = 6$. Therefore,

$$
S_{\mathrm{ABCD}} = \frac{1}{2} B D \cdot A C \sin 120^{\circ} = \frac{1}{2} \cdot 10 \cdot 6 \cdot \frac{\sqrt{3}}{2} = 15 \sqrt{3}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_5827e7e7a03e1b3b311bg-28.jpg?height=303&width=834&top_left_y=177&top_left_x=615)

Answer

$15 \sqrt{3}$.

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