2. Let $ABCD$ be a convex quadrilateral such that

$$
\measuredangle ABC=90^{\circ}, \measuredangle DAB=60^{\circ} \text{ and } \measuredangle BCD=120^{\circ}.
$$

Let the diagonals $AC$ and $BD$ intersect at point $M$, such that $\overline{BM}=1$ and $\overline{MD}=2$. Calculate the area of quadrilateral $ABCD$.
Solution. Given are the three angles in the quadrilateral $ABCD$, so it follows that $\measuredangle ADC=90^{\circ}$. Let $k$ be a circle with diameter $AC$. Since $\measuredangle ABC=90^{\circ}$ and $\measuredangle ADC=90^{\circ}$, it follows that $B, D \in k$, which means that $ABCD$ is a cyclic quadrilateral.

Triangle $BOD$ is isosceles with angle $\measuredangle DOB=120^{\circ}$ as a central angle, twice the corresponding peripheral angle $\measuredangle DAB=60^{\circ}$. It follows that $\Varangle OBD=30^{\circ}$, and from the right triangle $BOS$, where $S$ is the midpoint of $BD$, it is obtained that $\overline{OS}=\frac{\overline{OB}}{2}$. However, $\overline{BS}=\frac{\overline{BD}}{2}=\frac{\overline{BM}+\overline{MD}}{2}=\frac{3}{2}$, so from the Pythagorean theorem it follows: $\frac{\overline{OB}^{2}}{4}+\frac{9}{4}=\overline{OB}^{2}$, or $\overline{OB}=\sqrt{3}$ and $\overline{OS}=\frac{\sqrt{3}}{2}$.

Since $\overline{MS}=\overline{BS}-\overline{BM}=\frac{1}{2}$, from the right triangle $MOS$, it is obtained: $\overline{OM}=\sqrt{\overline{OS}^{2}+\overline{MS}^{2}}=\sqrt{\frac{3}{4}+\frac{1}{4}}=1$. It follows that $OBM$ is an isosceles triangle, from which it is obtained: $\measuredangle BOM=\measuredangle OBM=30^{\circ}$ and $\measuredangle OMB=120^{\circ}$. Triangle $ABO$ is isosceles, so

$$
2 \measuredangle BAM=180^{\circ}-\measuredangle AOB=180^{\circ}-(\measuredangle OBM+\measuredangle OMB)=180^{\circ}-150^{\circ}=30^{\circ}
$$

It follows that $\measuredangle BAM=15^{\circ}$ and $\measuredangle CAD=60^{\circ}-15^{\circ}=45^{\circ}$. Now, $\measuredangle ACD=45^{\circ}$, so $ACD$ is an isosceles right triangle. Its area is

$$
P_{ACD}=\frac{\overline{AD}^{2}}{2}=\frac{\overline{AC}^{2}}{4}=\frac{(2 \cdot \overline{AO})^{2}}{4}=\frac{(2 \cdot \overline{BO})^{2}}{4}=\frac{(2 \sqrt{3})^{2}}{4}=3
$$

If $BL$ is the height of triangle $ABC$ dropped to side $AC$, for triangle $BLM$ it is obtained that it is a right triangle with acute angles $30^{\circ}$ and $60^{\circ}$, so $\overline{ML}=\frac{1}{2}$ and $\overline{BL}=\frac{\sqrt{3}}{2}$. For the area of triangle $ABC$ it is obtained

$$
P_{ABC}=\frac{\overline{AC} \cdot \overline{BL}}{2}=\frac{2 \sqrt{3} \cdot \overline{BL}}{2}=\sqrt{3} \cdot \frac{\sqrt{3}}{2}=\frac{3}{2}
$$

Finally, the area $P$ of quadrilateral $ABCD$ is:

$$
P=P_{ABC}+P_{ACD}=\frac{3}{2}+3=\frac{9}{2}
$$