this is solvable

Line AB, CB, and DA are the same lengths. Angles 1 and 2 are the same.The angle of A is 120o .

The area of polygon ABCD is 60.

What is the area of the triangle ABD?

Let angles 1 and 2 are $\varphi$

Let AB, CB, and DA = a

1. $DB =\ ?$

$\begin{array}{|rcll|} \hline DB^2 &=& a^2+a^2-2\cdot a\cdot a\cdot \cos(120^{\circ}) \\ DB^2 &=& 2a^2[~1-\cos(120^{\circ}) ~] \quad & | \quad \cos(120^{\circ}) = \cos(180^{\circ}-60^{\circ}) = -\cos(60^{\circ}) \\ DB^2 &=& 2a^2[~1+\cos(60^{\circ}) ~] \quad & | \quad \cos(60^{\circ}) = \frac12 \\ DB^2 &=& 2a^2(~1+\frac12 ~) \\ DB^2 &=& 2a^2(~\frac32 ~) \\ DB^2 &=& 3a^2 \\ \mathbf{DB} & \mathbf{=} & \mathbf{a\cdot \sqrt{3}} \\ \hline \end{array}$

2. $\varphi =\ ?$

$\begin{array}{|rcll|} \hline \frac{ \sin(\varphi) } {a} &=& \frac{\sin(120^{\circ})} {DB} \\ \frac{ \sin(\varphi) } {a} &=& \frac{\sin(120^{\circ})} {a\cdot \sqrt{3}} \\ \sin(\varphi) &=& \frac{\sin(120^{\circ})} {\sqrt{3}} \quad & | \quad \sin(120^{\circ}) = \sin(180^{\circ}-60^{\circ}) = \sin(60^{\circ}) \\ \sin(\varphi) &=& \frac{\sin(60^{\circ})} {\sqrt{3}} \quad & | \quad \sin(60^{\circ}) = \frac{\sqrt{3}} {2} \\ \sin(\varphi) &=& \frac{\frac{\sqrt{3}} {2} } {\sqrt{3}} \\ \sin(\varphi) &=& \frac12\\ \mathbf{\varphi} & \mathbf{=} & \mathbf{30^{\circ}} \\ \hline \end{array}$

3. Angle $C =\ ?$

$\begin{array}{|rcll|} \hline \frac{ \sin(C) } {DB} &=& \frac{ \sin{30^{\circ}} } {a} \quad & | \quad \sin(30^{\circ}) = \frac12 \\ \frac{ \sin(C) } {a\cdot \sqrt{3}} &=& \frac{ \frac12 } {a} \\ \sin(C) &=& \frac{ \sqrt{3} } {2} \\ \mathbf{C} & \mathbf{=} & \mathbf{60^{\circ}} \\ \hline \end{array}$

So angle CBD is $90^{\circ}$ and triangle CBD is a right angular triangle.

4. The area of triangle CBD:

$\begin{array}{|rcll|} \hline A_{CBD} &=& \frac{ a\cdot DB }{2} \\ &=& \frac{ a\cdot a\cdot \sqrt{3} }{2} \\ \mathbf{A_{CBD}} & \mathbf{=} & \mathbf{\frac{ a^2\sqrt{3} }{2}} \\ \hline \end{array}$

5. The area of triangle ABD:

$\begin{array}{|rcll|} \hline A_{ABD} &=& \frac{ DB \cdot a \cdot \sin(\varphi) }{2} \\ &=& \frac{ a\cdot \sqrt{3} \cdot a \cdot \frac12 }{2} \\ A_{ABD} &=& \frac{ a^2\cdot \sqrt{3} }{4} \\ \text{or}\\ a^2 &=& \frac{4\cdot A_{ABD} } {\sqrt{3}} \\ \mathbf{a^2} & \mathbf{=} & \mathbf{\frac{4\cdot A_{ABD} } {\sqrt{3}}} \\ \hline \end{array}$

The area of polygon ABCD $A_{ABCD} = 60$

$\begin{array}{|rcll|} \hline A_{ABCD} &=& A_{ABD} + A_{CBD} \quad & | \quad A_{CBD} = \frac{ a^2\sqrt{3} }{2} \qquad A_{ABCD} = 60 \\ 60 &=& A_{ABD} + \frac{ a^2\sqrt{3} }{2} \quad & | \quad a^2 = \frac{4\cdot A_{ABD} } {\sqrt{3}} \\ 60 &=& A_{ABD} + \frac{ \frac{4\cdot A_{ABD} } {\sqrt{3}}\sqrt{3} }{2} \\ 60 &=& A_{ABD} + 2\cdot A_{ABD} \\ 60 &=& 3\cdot A_{ABD} \\ 20 &=& A_{ABD} \\ \mathbf{A_{ABD}} & \mathbf{=} & \mathbf{20} \\ \hline \end{array}$

The area of the triangle ABD is 20

$Area ABCD \\ = 2*ABD + BCE\\ = 2*(0.5x^2sin120) + 0.5x^2sin60\\ =2*(0.5 x^2sin60) + (0.5x^2sin60)\\ =3*(0.5 x^2sin60) \\ =3* area\;of\;\triangle ABD\\~\\ 60=3* area\;of\;\triangle ABD\\ Area\;of\;\triangle ABD=20\;\;u^2$

Like the way you did that one, Melody....!!!!