Dear student
 
There is no figure attached.
 
Please check and repost the question with an image.
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note that aruns ans is illogical, your ques is perfectly fine and doesnt need any pic. correct soln is mentioned below:
first draw the figure.
let BI when extended intersect AC at P and CI when extended intersect AB at Q. obviously X and Y lie on these lines respectively.
by properties of incentre, angle ABI= angle CBI= B/2
angle BCI= angle ACI= C/2
angle BAI= angle CAI= A/2.
now, in tri ABQ,
B/2 + A + angle AQB= 180
or angle AQB= 180 – (A+B/2)
now In tri AXQ,
90 + angle AQB + angle QAX= 180
substitute value of angle AQB, we have
angle QAX= A + B/2 – 90
Hence angle IAX= angle CAI – angle QAX
= A/2 – (A + B/2 – 90)
= 90 – (A+B)/2
= C/2
now, note that quad AYIX is a cyclic quad, since sum of opposite angles is 180 (angle AYI + angle AXI= 90+90= 180). hence a circle can be circumscribed around it. so, angle IYX= angle IAX (by angles in the same segment circle theorem).
so that angle IYX= angle IAX= C/2
but, angle BCI= C/2 too.
so, angle IYX= angle BCI
since angle IYX and angle BCI are alternate interior angles with the common transversal CY, we conclude that their equality implies that BC and XY are parallel.
KINDLY APPROVE :))