A geometry problem by Madhavarapu Revanth

Since $\dfrac {BD}{BC} = \dfrac {x_D-x_B}{x_C-x_B} = \dfrac {4-1}{5-1} = \dfrac {3}{4}$ , we find that $BD: DC = 3:1$ .

Let $\angle A$ , $\angle B$ and $\angle C$ be $\alpha$ , $\beta$ and $\gamma$ respectively. Using Sine Rule,

$\begin{cases} \dfrac {AB}{\sin{\left( 180^\circ-\frac{\alpha}{2}-\beta \right)}} = \dfrac {x}{\sin{\left( 180^\circ-\frac{\alpha}{2}-\beta \right)}} = \dfrac {3}{\sin{\frac{\alpha}{2}}} \\ \dfrac {AC}{\sin{\left( 180^\circ-\frac{\alpha}{2}-\gamma \right)}} = \dfrac {y}{\sin{\left( 180^\circ-\frac{\alpha}{2}-\gamma \right)}} = \dfrac {1}{\sin{\frac{\alpha}{2}}}\end{cases}$

But $\sin{\left( 180^\circ-\frac{\alpha}{2}-\beta \right)} = \sin{\left( 180^\circ-\frac{\alpha}{2}-\gamma \right)}$

$\Rightarrow x:y = 3:1\quad \Rightarrow x+y = \boxed{4}$