Geometry with circle and triangle. Any other solutions(advice) are welcome.
What I am asking is this because I am studying mathematics through feedback process of my solution and learning new solutions.
Please, release Hold on.
When will the hold be resolved? I want to see your new solutions(or advices).
$\overline{AE}$ is a diameter.
and $\overline{AB}=1,\;\;\; \overline{AC}=2$
and  $\overline{BD}:\overline{CD}=3:2$ 
Find the area of $\triangle ABC$ .



Here is a completely different soluiton based on trigonometry. 
From triangle $ABD$:
$$\frac{3x}{\sin\alpha}=\frac{l}{\sin(90^\circ-\beta)}\tag{1}$$
From triangle $ADC$:
$$\frac{2x}{\sin\beta}=\frac{l}{\sin(90^\circ-\alpha)}\tag{2}$$
From triangle $ABC$:
$$\frac{1}{\sin(90^\circ-\alpha)}=\frac{2}{\sin(90^\circ-\beta)}\tag{3}$$
Rewrite (1),(2)(3):
$$\frac{3x}{\sin\alpha}=\frac{l}{\cos\beta}\tag{4}$$
$$\frac{2x}{\sin\beta}=\frac{l}{\cos\alpha}\tag{5}$$
$$\cos\beta=2\cos\alpha\tag{6}$$
We don't care about $l$, just divide (4) by (5) to eliminate it:
$$\frac{3\sin\beta}{2\sin\alpha}=\frac{\cos\alpha}{\cos\beta}$$
$$3\sin\beta\cos\beta=2\sin\alpha\cos\alpha\tag{7}$$
Now you have (6) and (7), two equations with two unknown angles. Divide (7) by (6) and you get:
$$3\sin\beta=\sin\alpha\tag{8}$$
Now divide (6) by 2, square and add to (8) squared to eliminate $\alpha$:
$$9\sin^2\beta+\frac14\cos^2\beta=1$$
$$36\sin^2\beta+cos^2\beta=4$$
$$35\sin^2\beta=3$$
The rest is straightforward: 
$$\sin\beta=\frac{\sqrt{3}}{\sqrt{35}}, \ \ \cos\beta=\frac{4\sqrt{2}}{\sqrt{35}}$$
From (8):
$$\sin\alpha=\frac{3\sqrt{3}}{\sqrt{35}}, \ \ \cos\alpha=\frac{2\sqrt{2}}{\sqrt{35}}$$
Now:
$$S=\frac12 AB\cdot h=\frac12 \cdot 1 \cdot 2 \sin(\alpha+\beta)=\sin(\alpha+\beta)$$
$$S=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac{2\sqrt{6}}{5}$$
