Playing with Triangles!

Notice that $\angle B=\angle C$ , so let $\angle B=\angle C=\beta$ .

First we find $\sin \beta$ and $\cos \beta$ , this is easy since this is an isosceles triangle:

$\sin \beta=\dfrac{4}{5}$ and $\cos \beta=\dfrac{3}{5}$ .

Also, we find that $\cos \theta=\dfrac{12}{13}$ , $\sin(2\beta)=\dfrac{24}{25}$ and $\cos(2\beta)=-\dfrac{7}{25}$ . We are going to use these values later.

Then let $\angle BCF=\alpha$ , so $\angle BFC=180^\circ-\alpha-\beta$ and $\angle BEC=180-\beta-\theta$ . Now apply sine rule on $\triangle BFC$ :

$\dfrac{6}{\sin(180^\circ-\alpha-\beta)}=\dfrac{CF}{\sin \beta}$

And in $\triangle BEC$ :

$\dfrac{6}{\sin(180^\circ-\theta-\beta)}=\dfrac{BE}{\sin \beta}$

Since $BE=CF$ :

$\dfrac{6}{\sin(180^\circ-\alpha-\beta)}=\dfrac{6}{\sin(180^\circ-\theta-\beta)} \\ \sin(180^\circ-(\alpha+\beta))=\sin(180^\circ-(\theta+\beta))\\ \sin(\alpha+\beta)=\sin(\theta+\beta)$

But since $\alpha \neq \theta$ , then $\alpha=180^\circ-(\theta+2\beta)$ .

Now, $\angle BHC=180^\circ-\theta-\alpha=2\beta$ .

Then, calculate $BH$ using sine rule again:

$\dfrac{BH}{\sin \alpha}=\dfrac{6}{\sin \angle BHC} \\ BH=\dfrac{6\sin(180^\circ-(\theta+2\beta))}{\sin(2\beta)}=\dfrac{6\sin(\theta+2\beta)}{\sin(2\beta)}$

Finally,

$\sin \theta=\dfrac{HK}{BH} \implies HK=BH \sin \theta \\ HK=\dfrac{6\sin(\theta+2\beta) \sin \theta}{\sin(2\beta)} \\ HK=\dfrac{6(\sin \theta \cos(2\beta)+\cos \theta \sin(2\beta))\sin \theta}{\sin(2\beta)} \\ HK=\dfrac{6(\frac{5}{13}\cdot \frac{-7}{25}+\frac{12}{13}\cdot\frac{24}{25})(\frac{5}{13})}{\frac{24}{25}} \\ HK=\dfrac{1265}{676}$ .

Thus, the final answer is $1265+676=\boxed{1941}$ .

Let $BE=CF=a$ and $\angle ABC = \angle ACB = \phi = \sin^{-1} \left( \frac{4}{5} \right)$ . Using Sine Rule in $\triangle EBC$ , we have:

$\begin{aligned} \frac{a}{\sin{\phi}} & = \frac{6}{\sin{(180^\circ - \theta - \phi)}} = \frac{6}{\sin{(\theta + \phi)}} \\ \Rightarrow a & = \frac{6\sin{\phi}} {\sin{(\theta + \phi)}} = \frac{6\sin{\phi}} {\sin{\theta} \cos{\phi} + \sin{\phi} \cos{\theta}} \\ & = \frac{6\times \frac{4}{5}} {\frac{5}{13} \times \frac{3}{5} + \frac{4}{5} \times \frac{12}{13}} = \frac{104}{21} \end{aligned}$

Let $\angle BFC = \gamma$ . Using Sine Rule in $\triangle FBC$ , we have:

$\begin{aligned} \frac{\sin{\gamma}}{6} & = \frac{\sin{\phi}}{a} \\ \Rightarrow \sin{\gamma} & = 6 \times \frac{4}{5} \times \frac{21}{104} = \frac{63} {65} \quad \Rightarrow \cos{\gamma} = \frac{16}{65} \end{aligned}$

Then $\angle FCB = \delta = 180^\circ - \phi - \gamma$ .

$\begin{aligned} \Rightarrow \sin{\delta} & = \sin{(180^\circ - \phi - \gamma)} = \sin{(\phi + \gamma)} \\ & = \frac{4}{5}\times \frac{16}{65} + \frac{63}{65} \times \frac{3}{5} = \frac{253}{325} \\ \Rightarrow \tan{\delta} & = \frac{253}{204} \end{aligned}$

Now, let $BK = x$ . therefore, $KC = 6-x$ and $HK = y$ . Then, we have:

$\begin{aligned} y & = x \tan{\theta} = (6-x) \tan{\delta} \\ \Rightarrow \frac{5x}{12} & = \frac{253(6-x)}{204} \\ x & = \frac{759}{169} \\ \Rightarrow y & = x \tan{\theta} = \frac{759}{169} \times \frac{5}{12} = \frac{1265}{676} \\ \alpha + \beta & = 1265 + 676 = \boxed{1941} \end{aligned}$