I is the incenter of triangle ABC. X and Y are the feet of the perpendiculars from A to BI and CI. Proove that XY is parallel to BC.

Dear student

 

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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.