<s>[INST] Similar Triangles and Incircle The incircle of trianlge ABC touches the sides AB, AC and BC at points P, N, and M respectively. Denote AP=AN=x, BM=BP=y and CM=CN=z. Segment UV is tangent to the incircle at point X and parallel to the side AC. 
Prove $\cfrac{UV}{AC}$= $\cfrac{y}{x+y+z}$

What I have so far 
$\triangle$ ABC $\sim$ $\triangle$ BUV 
which has gotten $\cfrac{BV}{BC}$ = $\cfrac{BU}{AB}$ =$\cfrac{UV}{AC}$
any help would be appreciated! 
 [/INST] Let $a$, $b$, $c$ be the sidelengths of $BC$, $CA$, $AB$, let $h_1$ be the height from $B$ to $AC$, and let $h_2$ be the height from $B$ to $UV$.  Note that $\dfrac{UV}{AC} = \dfrac{h_2}{h_1}$ because of the similar triangles you mentioned.  Also, $h_2 = h_1 - 2r$ where $r$ is the inradius of triangle $ABC$.  Hence $$\dfrac{UV}{AC} = \dfrac{h_1-2r}{h_1} = 1-\dfrac{2r}{h_1}.$$  Now let $K$ be the area of triangle $ABC$.  We know $2K = h_1b = r(a+b+c)$, so $$\dfrac{r}{h_1} = \dfrac{2K/h_1}{2K/r} = \dfrac{b}{a+b+c}.$$ Thus $$\dfrac{UV}{AC} = 1-\dfrac{2b}{a+b+c} = \dfrac{a-b+c}{a+b+c} = \dfrac{y}{x+y+z}.\ \blacksquare$$  (Notes: The last step follows from $x = (b+c-a)/2, y=(c+a-b)/2, z=(a+b-c)/2$, which is well-known.  To get the formula $2K = r(a+b+c)$, note that $K$ is the sum of the areas of triangles $IAB$, $IBC$, and $ICA$.)
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