X and Y are points on the sides BC and AC of a triangle ABC respectively such that angle AXC = angle BYC and BX = XY. Prove that AX bisects the angle BAC.

Question

I believe the first step is to prove ABXY is a cyclic quadrilateral. Unfortunately, I don't know how!!!

Answer 1

let AX and BY intersect at D

since angle AXC = angle BYC
angle BXA = angle AYB (supplementary angles)

also since BX = XY
then
angle XBY = XYB (isosceles triangles) mark that angle o

now in triangles BXD and ADY you have opposite angles equal and angle BXA = angle AYB (see above)

so the third angle pair must be equal (supplementary angles)

then angle XBD = angle YAX = o

so now you have equal angles XBD and YAX subtended by XY
Therefore ABXY is a cyclic quadrilateral (properties of a cyclic quad)

then by the same cyclic quad properties
angles BAX and BYX are subtended by the same side BX in that cyclic quad.
Therefore angles BAX and BYX are equal, marked as o

QED !