Centroid, Incentre, Orthocentre?

We shall use Ceva's theorem to convert the concurrency condition into an equivalent condition involving ratio of side lengths.

Referring to the figure, Ceva's theorem gives us $\frac{AF}{FC} \cdot \frac{CD}{DB} \cdot \frac{BE}{EA}=1$

Using the angle bisector theorem we have that $\frac{AF}{FC} = \cos \alpha$

Using the definition of a median , we have $\frac{BE}{EA} =1$

Now, using a little trigonometry, we get the folowing:

$\begin{aligned} \frac{AB}{BC} &= \cos \alpha \\ \frac{BD}{AB} &= \cos \alpha \\ \implies BD &= BC \cos^2 \alpha \\\implies CD &= BC \sin^2 \alpha \\\implies \frac{CD}{DB} &= \tan^2 \alpha \end{aligned}$

So, plugging all these into the first equation, we have:

$\begin{aligned} \cos \alpha \tan^2 \alpha &= 1 \\\implies \tan^2 \alpha &= \sec \alpha \\\implies \sec^2 \alpha &= \sec \alpha + 1 \\\implies \sec \alpha &= \varphi \quad \left(\because \sec \alpha \geq 1 \right) \end{aligned}$

Where $\varphi$ is the golden ratio .

Now, $\begin{aligned} \cos \alpha &= \frac{1}{\varphi} \\\implies \cos^2 \alpha &= \frac{1}{\varphi^2} \\ &= \frac{1}{1+\varphi} \\ &= \frac{1}{1+\frac{1+\sqrt5}{2}} \\ &= \frac{2}{3+\sqrt5} \end{aligned}$

Therefore, $a=2 \\ b=3 \\ c=5 \\ a+b+c= \boxed{10}$

Sorry Mr Deeparaj Bhat , I don't review your solution before submitting my private.

Relevant wiki: 2D Coordinate Geometry - Problem Solving

Let $A=(0,0), B=(0,2p), C=(2q,0)$ and $F=(r,0)$ . Then $E=(0,p)$ .

Now the gradient of $BC$ is $-\frac{p}{q}$ and the gradient of $CE$ is $-\frac{p}{2q}$ . This means that the equation of $AD$ is $py=qx$ and the equation of $CE$ is $y=-\frac{px}{2q}+p$ . Solving these equation, we get the intersect point $G=\left(\frac{2p^2q}{p^2+2q^2},\frac{2pq^2}{p^2+2q^2}\right)$ .

Next, since $BF$ bisects $\angle ABC$ , $\frac{2p}{2\sqrt{p^2+q^2}}=\frac{r}{2q-r}$ , which implies that $r=\frac{2pq}{p+\sqrt{p^2+q^2}}$ .

Since the points $B, G$ and $F$ collinear, the gradient of $BF$ equals to the gradient of $BG$ , which means that $\frac{-2p}{\frac{2pq}{p+\sqrt{p^2+q^2}}} = \frac{\frac{2pq^2}{p^2+2q^2}-2p}{\frac{2p^2q}{p^2+2q^2}}$ . After manipulation, we obtained $q^2=\frac{1+\sqrt{5}}{2}p^2$ .

Finally, $\cos^2\alpha = \frac{p^2}{p^2+q^2}=\frac{p^2}{p^2+\frac{1+\sqrt{5}}{2}p^2}=\frac{2}{3+\sqrt{5}}=\frac{a}{b+\sqrt{c}}$ . So $a+b+c=\boxed{10}$ , as $c$ is prime.