In a convex quadrilateral $A B C D$, the bisector of angle $A B C$ intersects side $A D$ at point $M$, and the perpendicular dropped from vertex $A$ to side $B C$ intersects $B C$ at point $N$, such that $B N=N C$ and $A M=2 M D$. Find the sides and area of quadrilateral $A B C D$, if its perimeter is $5+\sqrt{3}$, and angle $B A D$ is $90^{\circ}$ and angle $A B C$ is $60^{\circ}$.
Quadrilateral $ABCD$ is a trapezoid.

## Solution

Let $F$ and $Q$ be the projections of point $C$ onto lines $AB$ and $AD$ respectively. Denote $BN = NC = x$. Then

\[
\begin{gathered}
AB = 2x, AM = \frac{2x \sqrt{3}}{3}, MD = \frac{x \sqrt{3}}{3} \\
AQ = FC = BC \cdot \sin 60^{\circ} = x \sqrt{3}
\end{gathered}
\]

i.e., $AQ = AD$. Therefore,

\[
CD = AF = AB - BF = 2x - x = x
\]

According to the problem,

\[
AB + BC + CD + AD = 5 + \sqrt{3} \text{, or } 2x + 2x + x + \sqrt{3}x = (5 + \sqrt{3})x = 5 + \sqrt{3}
\]

From this, we find that $x = 1$.

Since $ABCD$ is a trapezoid,

\[
S_{\mathrm{ABCD}} = \frac{AB + CD}{2} \cdot AD = \frac{2 + 1}{2} \cdot \sqrt{3} = \frac{3 \sqrt{3}}{2}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-26.jpg?height=395&width=529&top_left_y=1852&top_left_x=770)

## Answer

$AB = BC = 2, AD = \sqrt{3}, DC = 1, S_{\mathrm{ABCD}} = \frac{3 \sqrt{3}}{2}$.