parmenides51 wrote: In triangle $ABC$, $\angle BAC= 60^o$. Angle bisector of $\angle ABC$ meets side $AC$ at $X$ and angle bisector of $\angle BCA$ meets side $AB$ at $Y$. Prove that if $I$ is the incenter of triangle $ABC$, then $IX=IY$. Note that $\angle XIY=120^\circ$ so $A,X,I,Y$ is concyclic. But we have $\angle YAI=\angle XAI$ too, so $IX=IY$.