Given `Delta ABC` with AB=3,BC=2,AC=4. A tangent to the incircle is drawn parallel to `bar(BC)` intersecting `bar(AB)` at X and `bar(AC)` at Y. Find the length of `bar(XY)` .

The radius of the incenter can be found by `r=(2a)/p` where a is the area of the triangle, and p is the perimeter of the triangle.

Using Heron's formula we find `a=sqrt(9/2(5/2)(3/2)(1/2))=(3sqrt(15))/4` and p=9.

So `r=((3sqrt(15))/2)/9=sqrt(15)/6`

The area of the triangle is also given by `a=1/2bh` where b is the length of a base and h the length of the associated height. Let b=2, then `a=1/2(2)h=h=(3sqrt(15))/4` .

The diameter of the incircle is `2r=sqrt(15)/3` . The ratio of the diameter of the incircle to the height of `Delta ABC` is `(sqrt(15)/3)/((3sqrt(15))/4)=4/9` . The height of `Delta AXY` will have a ratio to the height of `Delta ABC` of 5/9.

`Delta AXY ` ~ `Delta ABC` and the scale factor is 5/9.

Then `(XY)/2=5/9 ==> XY=10/9` so the correct answer is B.

** Note that X and Y are not tangent to the incircle. AX and AY are in a ratio of 3:4. In fact, AX=5/3,XB=4/3,AY=20/9,YC=16/9.

You need to remeber that if the sides of lengths 3 and 4 are cut at X and Y (tangency points), hence, the following relations are verified:

`CX=CY = (3+4-2)/2 = 5/2`

You need to draw the parallel line XY to the length of 2, hence, the following relations in similar triangles `Delta CXY ` and `Delta CAB` are verified:

`(CX)/3 = (CY)/4 = (XY)/2`

Substituting `5/2 ` for `CX ` yields:

`(5/2)/3 = (XY)/2 => 5/6 = (XY)/2 => XY = 5/3`

**Hence, evaluating the length of the parallel line XY yields `XY =
5/3` , hence, you need to select the answer `A. 5/3` .**

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