Find an angle $\angle BXY$ in a given triangle $\triangle ABC$

Question

In $\triangle ABC$, the lines $BP$, $BQ$ trisect $\angle ABC$ and the lines $CM$, $CN$ trisect $\angle ACB$. Let $BP$ and $CM$ intersect at $X$ and $BQ$ and $CN$ intersect at $Y$. If $\angle ABC=45^\circ$ and $\angle ACB=75^\circ$, then the angle $\angle BXY$ is $$(a)45^\circ\quad (b)47.5^\circ\quad (c)50^\circ\quad (d)55^\circ.$$ After some calculation I found the following relation: $$\angle XSY=\angle XTY-10^\circ.$$ But I can not go further. Give me some hints.

Answer 1

In the $BXC$ triangle the bisector that leaves the vertex $X$ must pass through the intersection point of the bisectors that leave $B$ and $C$, which is not other than $Y$. Consequently the requested angle is $50^{\circ}$ (that angle $BXC$ is $100^{\circ}$ is too easy to find out).

Answer 2

HINTS

Sum of angles in triangle = $180^0$

Sum of angles in straight line = $180^0$

Sum of angles in quadrilateral = $360^0$.

Your shape is made up of a lot of quadrilaterals and triangles, use these rules to work your way around the edges and in.

So far I've got that $SXY+TXY=100^0$, and $SYX+TYX=140^0$, and $SXY+SYX=125^0$ and $TXY+TYX=115^0$

I'm not certain (but have a gut feeling), that you can cut assume $XY$ bisects $SXT$, but if you can, then your answer is $50^0$

Edit: Due to Piquito's answer, it would appear that you can.