<s>[INST] Compute the area of Quadrilateral $ABCD$ As title suggests, the question is to solve for the area of the given convex quadrilateral, with two equal sides, a side length of 2 units and some angles:

I have solved the problem with a synthetic geometric approach involving some angle chasing. However, I believe my solution (which I will post in a comment below since I don't want to clutter the question) is a little messy and not efficient. Are there any better ways to do this? Geometric and/or trigonometric approaches are all welcomed!
EDIT: I have posted my solution below!
 [/INST] Alternative approach:
Area$(\triangle ADC) = \dfrac{1}{2} \times \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)].$
Area$(\triangle ABC) = \dfrac{1}{2} \times \overline{BC} \times [ ~\overline{AC} \sin(75^\circ)].$
Therefore, $~\text{Area}(\triangle ADC) ~=~ \text{Area}(\triangle ABC).$
Therefore
$$\text{Area(quadrilateral)} ~=~ \overline{AD} \times [ ~\overline{AC} \sin(105^\circ)]. \tag1 $$
Plan of Attack:

*

*Let $X$ denote the intersection of the two diagonals.


*Use the
Law of Sines
and the
Law of Cosines to compute
$~\overline{AX}, ~\overline{BX}, ~\overline{CX}, ~\overline{DX}.$


*Compute $~\overline{AD}.$


*Apply the formula in (1) above.

$\displaystyle \frac{\overline{DX}}{\sin(105^\circ)} = 
\frac{\overline{AD}}{\sin(45^\circ)} = 
\frac{\overline{BC}}{\sin(45^\circ)} = 
\frac{\overline{BX}}{\sin(75^\circ)} \implies $
$$\overline{DX} = \overline{BX}. \tag2 $$
$\displaystyle \frac{\overline{AX}}{\sin(30^\circ)} = 
\frac{\overline{AD}}{\sin(45^\circ)} = 
\frac{\overline{BC}}{\sin(45^\circ)} = 
\frac{\overline{CX}}{\sin(60^\circ)} \implies $
$\displaystyle \frac{1}{2} \times \overline{CX} = \frac{\sqrt{3}}{{2}} \times \overline{AX} \implies $
$$\overline{CX} = \sqrt{3} \times \overline{AX}. \tag3 $$

$$\overline{DX}^2 + \overline{AX}^2 
- \left[2\overline{AX}~~\overline{DX} \times \frac{1}{\sqrt{2}}\right]$$
$$= ~\overline{CX}^2 + \overline{BX}^2 
- \left[2\overline{CX}~~\overline{BX} \times \frac{1}{\sqrt{2}}\right] \implies $$
$$\overline{DX}^2 + \overline{AX}^2 
- \left[\sqrt{2} ~\overline{AX}~~\overline{DX}\right]$$
$$= ~3\overline{AX}^2 + \overline{DX}^2 
- \left[\sqrt{2} ~\left(\sqrt{3} ~\overline{AX}\right)~~\overline{DX}\right] \implies $$
$$= ~2\overline{AX}^2 + \left[ ~\sqrt{2} ~\overline{AX}~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies $$
$$= ~2\overline{AX} + \left[ ~\sqrt{2} ~~\overline{DX} ~~\left(1 - \sqrt{3}\right) ~\right] = 0 \implies $$
$$\overline{AX} ~=~ \overline{DX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right] 
~=~ \overline{BX} ~~\left[\frac{\sqrt{3} - 1}{\sqrt{2}}\right]. \tag4 $$

Note $\displaystyle ~: ~\cos(135^\circ) = \frac{-1}{\sqrt{2}}.$
\begin{alignat*}{2}
4 ~ &=~ \overline{AX}^2 + \overline{BX}^2  + \sqrt{2} ~\overline{AX} ~~\overline{BX} \\ \\ 
&=~ \overline{BX}^2\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right)^2
~+~ \overline{BX}^2 
~+~ \sqrt{2} ~\left(\frac{\sqrt{3} - 1}{\sqrt{2}}\right) ~\overline{BX}^2 \\ \\ 
&=~ \overline{BX}^2 ~\left[ ~\left( ~\frac{4 - 2\sqrt{3}}{2} ~\right) ~+~ 1 ~+~ \sqrt{3} - 1 ~\right] \\ \\ 
&=~ \overline{BX}^2 ~\left[ ~\frac{4 - 2\sqrt{3}}{2} ~+~ \frac{2\sqrt{3}}{2} ~\right] \\ \\ 
&=~ 2 ~\overline{BX}^2 \implies 
\end{alignat*}
$$\overline{BX} = \sqrt{2}. \tag5 $$

Therefore

*

*$~\overline{AX} = \sqrt{3} - 1.$


*$~\overline{BX} = \sqrt{2}.$


*$~\overline{CX} = 3 - \sqrt{3}.$


*$~\overline{DX} = \sqrt{2}.$

$\underline{\text{Computation of} ~\overline{AD}}$
\begin{alignat*}{2}
\overline{AD}^2 ~ &=~ \overline{AX}^2 + \overline{DX}^2  - \sqrt{2} ~\overline{AX} ~~\overline{DX} \\ \\ 
&=~ 4 - 2\sqrt{3} + 2 - 2\sqrt{3} + 2 \\ \\
&= 8 - 4\sqrt{3} \implies 
\end{alignat*}
$$\overline{AD} = \sqrt{8 - 4\sqrt{3}}. \tag6 $$

$\underline{\text{Final Computations}}$
In general,
$\cos(2\theta) = 2\cos^2(\theta) - 1 \implies $
$\displaystyle \frac{\sqrt{3}}{2} = \cos(30^\circ) = 2\cos^2\left(15^\circ\right) - 1 \implies $
$\displaystyle \cos\left(15^\circ\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \frac{1}{2} \sqrt{2 + \sqrt{3}} = 
\sin\left(75^\circ\right) = \sin\left(105^\circ\right).$
$\overline{AC} = \overline{AX} + \overline{CX} = 2.$
Therefore, using (1) above,
$$\text{Area(quadrilateral)} ~=~ 
\sqrt{8 - 4\sqrt{3}} \times 2 \times 
\frac{1}{2} \sqrt{2 + \sqrt{3}}$$
$$=~ 2 \times \sqrt{2 - \sqrt{3}} \times \sqrt{2 + \sqrt{3}} = 2.$$
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