## Task A-3.7.

Let $A B C D$ be a tangential quadrilateral in which $\varangle D A B=\varangle A B C=120^{\circ}$ and $\varangle C D A=90^{\circ}$. If $|A B|=1$, calculate the perimeter of quadrilateral $A B C D$.
## First Solution.

The fourth angle of the given quadrilateral is $\varangle B C D=30^{\circ}$. 1 point

Let $E$ be the intersection of lines $B C$ and $D A$.

Let $|D E|=x$.

![](https://cdn.mathpix.com/cropped/2024_05_30_abcf4bcd4243d7db1a46g-17.jpg?height=496&width=1025&top_left_y=740&top_left_x=447)

The measures of angles $\varangle A B E$ and $\varangle B A E$ are $60^{\circ}$, so triangle $A B E$ is equilateral, and thus $|A E|=|B E|=|A B|=1$.

Triangle $C D E$ is a right triangle, with acute angles $\varangle D C E=30^{\circ}$ and $\varangle C E D=60^{\circ}$. Therefore, $|C E|=2 x$ and $|C D|=x \sqrt{3}$.

Now, let's determine the lengths of all sides of quadrilateral $A B C D$:

$$
|A B|=1, \quad|B C|=2 x-1, \quad|C D|=x \sqrt{3}, \quad|D A|=x-1
$$

Since quadrilateral $A B C D$ is tangential, we have $|A B|+|C D|=|B C|+|D A|$, i.e.,

$$
\begin{gathered}
1+x \sqrt{3}=(2 x-1)+(x-1) \\
(3-\sqrt{3}) x=3
\end{gathered}
$$

Thus, $x=\frac{3}{3-\sqrt{3}}=\frac{3+\sqrt{3}}{2}$.

The perimeter of quadrilateral $A B C D$ is

$$
\begin{aligned}
|A B|+|B C|+|C D|+|D A| & =1+(2 x-1)+x \sqrt{3}+(x-1) \\
& =(3+\sqrt{3}) x-1 \\
& =\frac{1}{2}(3+\sqrt{3})^{2}-1 \\
& =5+3 \sqrt{3}
\end{aligned}
$$