Question 31170
LET AB AND CD BE THE PARALLEL LINES.LET XY BE A TRANSVERSAL CUTTING AB AT P AND CD AT Q.LET EF BE THE BISECTOR OF ANGLE XPB AND GH THE BISECTOR OF ITS CORRESPONDING ANGLE XQD.
CORRESPONDING ANGLES ARE EQUAL SO..
ANGLE XPB=ANGLE XQD....SINCE EF AND GH ARE THEIR BISECTORS....WE HAVE ....
ANGLE XPF = ANGLE XPB/2 = ANGLE XQD/2 = ANGLE XQH....SO.....
WE HAVE 2 LINES EF AND GH ,A TRANSVERSAL XPQY CUTS THEM AT P AND Q RESPECTIVELY.
AND....ANGLE XPF = ANGLE XQH...WHICH ARE CORRESPONDING ANGLES . HENCE EF IS PARALLEL TO GH.SO THE BISECTORS OF CORRESPONDING ANGLES ARE PARALLEL.