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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.

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8 months ago

```							Dear student There is no figure attached. Please check and repost the question with an image.I will be happy to help you.
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8 months ago
```							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/2angle BCI= angle ACI= C/2angle BAI= angle CAI= A/2.now, in tri ABQ,B/2 + A + angle AQB= 180or angle AQB= 180 – (A+B/2)now In tri AXQ,90 + angle AQB + angle QAX= 180substitute value of angle AQB, we haveangle QAX= A + B/2 – 90Hence angle IAX= angle CAI – angle QAX= A/2 – (A + B/2 – 90)= 90 – (A+B)/2= C/2now, 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/2but, angle BCI= C/2 too.so, angle IYX= angle BCIsince 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 :))
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8 months ago
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