Let $I_1$ be incenter of $\triangle AA_1C$ and $I_2$ be incenter of $\triangle AC_1C$ Hence, $Y-I_2-I_1-X$ Lines $AI_2$ and $CI_1$ meet at $I$ which is the incenter of $\triangle ABC$ Let $D$ be on $BC$ such that $ID \perp BC$ $\angle AI_2C = 90 + \frac{\angle AC_1C}{2} = 135$ Similarly, $\angle AI_1C = 135$ Therefore, quadrilateral $AI_2I_1C$ is cyclic Hence, $\angle II_2I_1 = \angle I_1CA = \frac{C}{2}$ Therefore, $\angle II_2X = \frac{C}{2}$ $\angle IDX = 90$ $\angle I_2ID = \angle AIB + \angle BID = 90 + \frac{C}{2} + 90 - \frac{B}{2}$ Therefore, in quadrilateral $I_2IDX$, reflex $\angle I_2ID = 180 + \frac{B}{2} - \frac{C}{2}$ Therefore, By angle sum property of quadrilateral, $\angle DXI_2 = 90 - \frac{B}{2}$ Hence, $\angle BXY = 90 - \frac{B}{2}$ In $\triangle BXY$, $\angle YBX = B$ and $\angle BXY = 90 - \frac{B}{2}$ Hence, $\angle BYX = 90 - \frac{B}{2} \implies \angle BXY = \angle BYX \implies BX=BY$

Cute parmenides51 wrote: Let $ AA_1 $ and $ CC_1 $ be the altitudes of the acute-angled triangle $ ABC $. A line passing through the centers of the inscribed circles the triangles $ AA_1C $ and $ CC_1A $ intersect the sides of $ AB $ and $ BC $ triangle $ ABC $ at points $ X $ and $ Y $. Prove that $ BX = BY $. Note that $\angle AI_1C=\angle AI_2C=135^{\circ}$ so $A, I_1, I_2, C$ are concyclic. Since $IB$ passes through the circumcentre of $\triangle IAC$, it is perpendicular to line $I_1I_2$, hence $BX=BY$.