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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 $$(a)45^\circ\quad (b)47.5^\circ\quad (c)50^\circ\quad (d)55^\circ.$$ enter image description here After some calculation I found the following relation: $$\angle XSY=\angle XTY-10^\circ.$$ But I can not go further. Give me some hints.

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  • $\begingroup$ $\angle XSY \neq \angle XTY$ $\endgroup$ – Michael Biro May 9 '18 at 12:57
  • $\begingroup$ @Michael Biro Sorry.... My mistake. Thanks for mentioned it. $\endgroup$ – SAHEB PAL May 9 '18 at 13:02
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    $\begingroup$ I think you'll have to assume a side length of one side of the triangle (BC =10) so you can figure length BX and BY and then angle BXY using trig. I don't think you can do it with angles alone. It doesn't matter what length you chose as all similar triangles have the same angles. $\endgroup$ – Phil H May 9 '18 at 13:38
  • $\begingroup$ @PhilH. Not necessary actually, the three basic angle principles will get you through it all $\endgroup$ – Rhys Hughes May 9 '18 at 14:18
  • $\begingroup$ I figured it out with my method which came out to 50 deg; so with a result which is 1/2 of angle BXC, there must be an easier way. Thanks $\endgroup$ – Phil H May 9 '18 at 18:00
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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).

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  • $\begingroup$ Very nice observation! $\endgroup$ – Michael Biro May 9 '18 at 14:27
  • $\begingroup$ @Michael Biro: Thank you very much for the compliment. "Observation" is, in effect, the key to many elementary problems in Euclidean geometry. $\endgroup$ – Piquito May 9 '18 at 14:53
  • $\begingroup$ $Y$ is thus the incenter of $BXC$ $\endgroup$ – Dan Uznanski May 9 '18 at 15:10
  • $\begingroup$ @Dan Uznanski: So is. But what is most important here is that the inner bisectors of a triangle concur at a point. $\endgroup$ – Piquito May 9 '18 at 15:19
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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.

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  • $\begingroup$ Using trig form of civa's theorem you can prove that XY is the bisector $\endgroup$ – Thishanka Alahakoon May 10 '18 at 17:08

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