Question 762959

See the picture below which describes your problem. I have extended line CA further upwards to a point P, which is required for the proof.

{{{drawing (600,600,-10,10,-20,20, 
line(-2,3.5,8,3.5),
line(-1,9,8,9),
line(-2,3.5,-0.55,12),
line(-2,3.5,0.5,6.5),
line(-1,9,0.5,6,5),
locate(0.5,6.5,B),
locate(-1,9,A),
locate(8,9,X),
locate(-2,3.5,C),
locate(8,3.5,Y),
locate(-0.55,12,P)
)
}}}

Problem:
AX || CY
BA bisects ?CAX,
BC bisects ?ACY.
Prove: ?B is a right angle. 

Step 1:
BA bisects angle CAX. So angle CAB = angle BAX

Step 2:
BC bisects angle ACY. So angle ACB = angle BCY

Step 3:
Angle PAX = angle ACY since AX || CY (external angles of parallel lines)

Step 4:
Angles PAX + CAX = 180 (supplementary angles on a straight line)

Step 5:
i.e. PAX / 2 + CAX / 2 = 90

Step 6:
ACY / 2 + CAX / 2 = 90  Since PAX = ACY

Step 7:
ACB + CAB = 90  Since ACB is half of ACY, CAB is half of CAX

Step 8:
But in Triangle ABC, ACB + CAB + ABC = 180

Step 9:
So ABC = Angle B = 90

:)