Deriving formula from geometric relations If we know that 
and we have this triangle

Then how can we derive this relation? 
($\alpha$ and $\beta$ are assumed to be small, so that $\tan(\alpha) \approx \alpha$ and $\tan(\beta) \approx \beta$). We also know that $\theta = \alpha + \beta$, so
$$\alpha \approx \frac{4GM}{c^2b}-\tan(\beta)=\frac{4GM}{c^2b}-\frac{b}{D}$$
Also, $b=d\tan(\alpha)=D\tan(\beta)$. But what next? Where is the square root coming from?
 A: $$\tan \alpha=\frac{b}{d} \text{ and } \tan \beta=\frac{b}{D}$$
so,
$$\frac{D/d}{D+d}=\frac{1}{b}\frac{\tan \alpha/\tan \beta}{\frac{1}{\tan \alpha}+\frac{1}{\tan \beta}}=\frac{1}{b}\frac{\tan^2 \alpha}{\tan \alpha+\tan \beta}$$
and,
$$\frac{4GM}{c^2}\frac{D/d}{D+d}=\theta\cdot b\cdot \frac{1}{b}\frac{\tan^2 \alpha}{\tan \alpha+\tan \beta}=\frac{\theta\cdot\tan^2 \alpha}{\tan \alpha+\tan \beta} \quad (*)$$
Using that $\tan\alpha \approx \alpha$ and $\tan\beta \approx \beta$ we get
$$\frac{\theta}{\tan \alpha+\tan \beta}\approx \frac{\theta}{ \alpha+ \beta}=1$$
and backing to $(*)$ we get
$$\alpha^2=\frac{4GM}{c^2}\frac{D/d}{D+d}\to \alpha=\sqrt{\frac{4GM}{c^2}\frac{D/d}{D+d}}$$
A: I have found another solution following your idea. Take your last expression:
$$\alpha =\frac{4GM}{c^2b}-\frac{b}{D}$$
Now use that $b=d\tan \alpha\approx d\alpha$:
$$\alpha =\frac{4GM}{c^2d\alpha}-\frac{d\alpha}{D}\to \alpha^2=\frac{4GM}{c^2d}-\frac{d\alpha^2}{D}\\
\alpha^2+\frac{d\alpha^2}{D}=\frac{4GM}{c^2d}\to \alpha^2 \left(1+\frac{d}{D}\right)=\frac{4GM}{c^2d}\\
\alpha ^2=\frac{4GM}{c^2d}\cdot \frac{D}{d+D}\to \alpha =\sqrt{\frac{4GM}{c^2d}\cdot \frac{D}{d+D}}\to\\
\alpha =\sqrt{\frac{4GM}{c^2}\cdot \frac{D/d}{d+D}}$$
