How to solve this semi-circle problem? 

The figure above shows a semi-circle. $\angle BAP$ is $\alpha$ radian. The area of semi-circle is bisected by $AP$. Prove that $$2\alpha+\sin 2\alpha = \frac{\pi}{2}$$

I have simply no clue to solve this problem. Can anyone help me? Thanks so much.
 A: Area of semi-circle $= \frac{1}{2} \pi r^{2}$.
Let $O$ be the centre.  Then
$\angle BOP = 2\alpha$ and $\angle AOP = \pi-2\alpha$
Area of $ABP=$ Area of sector $OBP +$ Area of $\Delta OAP$
$\implies \frac{1}{4} \pi r^{2} =
\frac{1}{2} r^{2}(2\alpha)+\frac{1}{2} r^{2} \sin (\pi-2\alpha)$
$\implies \frac{1}{4} \pi r^{2} = r^{2}\alpha+\frac{1}{2} r^{2} \sin 2\alpha$
$\implies \frac{1}{2} \pi = 2\alpha+ \sin 2\alpha$
(QED)
Note that the radius $r$ will be automatically cancelled out.
P.S.: The word "bisect" is not good.  May say "$AP$ divides the semi-circle into $2$ equal areas".
A: Let $r$ be the radius & $O$ be the center of the semi-circle.
Area of the segment ABP of semi-circle 
$$(\text{area of sector OBP with angle of aperture } 2\alpha)+(\text{area of isosceles}\ \triangle AOP)$$
$$=\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(\pi-2\alpha)=\color{blue}{\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(2\alpha)}$$
But the area of semi-circle is bisected by $AP$ hence, Area of the segment ABP of semi-circle 
$$=\text{half the area of semi-circle}=\frac 12\frac{\pi r^2}{2}=\color{blue}{\frac{\pi r^2}{4}}$$
Equating both the values, $$\frac{1}{2}r^2(2\alpha)+\frac{1}{2}r^2\sin(2\alpha)=\frac{\pi r^2}{4}$$
$$\color{red}{2\alpha+\sin(2\alpha)=\frac{\pi}{2}}$$
