How to finish the following proof? The incircle of $\triangle ABC$ touches the sides $AB, AC,$ and $BC$ at points $P, N, M$, respectively. Denote $AP = AN = x, BM = BP = y, CM = CN = z$, as tangents from an exterior point to a circle are congruent. Segment $UV$ is tangent to the incircle and parallel to the side $AC$. 
Prove $\displaystyle\frac{UV}{AC} = \frac{y}{x+y+z}$ 
So far I have: Since $\triangle BUV \sim \triangle BAC$ $\rightarrow$  $\displaystyle\frac{UV}{AC} = \frac{BU}{BA} = \frac{BV}{BC} = \frac{BU}{BA} \cdot \frac{BC}{BV}$. 
I tried plugging in the values for the sides and don't end up with the relationship that is needed to prove. Any and all help is appreciated. 
 A: First of all, let's call $R$ the intersection of $UV$ and the circle. 
So, $PU=UR$ and $MV=VR$ so the perimeter if the triangle $PBM$ is 
$$BU+UV+VB=(BP-PU)+(UR+VR)+(BM-MV)=BP+BM=2BP=2y$$ 
once $UV$ is parallel to $AC$ then the triangles $BUV$ and $BAC$ are similiars. So,
$$\frac{UV}{AC}=\frac{\text{perimeter}(BUV)}{\text{perimeter}(BAC)}=\frac{2y}{2(x+y+z)}=\frac{y}{x+y+z}$$
A: The altitude of $\triangle ABC $ passing through $B$ has length which may be calculated by,$$rs = \frac{1}2 AC \cdot BN \ \ \ \Rightarrow \ \ \ BN = \frac{2rs }{AC} = \frac{2r(x+y+z)}{x+z} \tag{1}$$ where $s$ denotes the semiperimeter of $\triangle ABC$. (This is because the area of a triangle is $rs$ where $r$ is the inradius). And then let's calculate the altitude of the smaller triangle passing through $B$ again. We know from the figure that it's equal to$$BN - 2r = 2r \left(\frac{x + y +z}{x+z} - 1\right) = \frac{2 ry}{x+z}\tag{2}$$And then using the fact that these two triangles are similar, which you observed, the ratio of the two sides is also the ratio of altitudes, and they are given in $(1)$ and $(2)$. You get your result as soon as you divide them.
