Line BL belongs to triangle BCL
Line CM belons to triangle BCM
The triangles BCL and BCM has a coomon side, the side BC, this side is congruent.
If Angle XBA = Angle YCA, then:
Angle ABC or angle MBC of triangle BCM = Angle ACB or angle LCB of triangle BCL
If line BE is the bisector of angle ABC, then divides it into two equal parts, then angle MBL must be congruent with angle LBC
If line CD is the bisector of angle ACB, then divides it into two equal parts, then angle LCM must be congruent with angle MCB
As angle MBC is congruent with angle LCB, the angles MBL, LBC, LCM, and MCB must be congruents too.
Then angle MCB in triangle MBC is congruent with angle LBC of triangle BCL.
Then, the two triangles BCM and BCL have a congruent side (BC) and the two adjacent angles are congruent too (angle MBC of triangle BCM with angle LCB of triangle BCL, and angle MCB of triangle BCM with angle LBC of triangle BCL).
Then by ASA the two triangles are congruents and the other two sides must be congruents too: Line BL of triangle BCL must be congruent with line CM of triangle BCM (because they are opposite to congruent angles: BL in triangle BCL is opposite to angle LCB, and CM in triangle BCM is opposite to angle MBC, and angles LCB and MBC are congruents).
Answer: I would use ASA to help me prove Line BL= Line CM
