# 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!

• Do you mean $AP = AN = x$? – gt6989b May 19 '17 at 19:09
• @gt6989b Yes! sorry! – Parley May 19 '17 at 19:10

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 $ICA$.)

• how did u get $\cfrac{UV}{AC}$=$\cfrac{h_{2}}{h_{1}}$ – Parley May 19 '17 at 21:23
• Also, where did the whole "notes" section come from? what is x, y and z? – Parley May 19 '17 at 21:37
• $UV/AC = h_2/h_1$ comes from the similar triangles $ABC$ and $UBV$. $x$, $y$, and $z$ are the same as they are in the diagram in your question. Hope that helps! – fractal1729 May 20 '17 at 0:33
• Yes! can u just explain how u arrived at x y and z – Parley May 20 '17 at 0:42
• $y+z = a$, $z+x = b$, $x+y = c$ is a system of three equations in the three variables $x,y,z$. Solving the system in terms of $a,b,c$ yields the three formulas for $x,y,z$ that I mentioned in my post. – fractal1729 May 20 '17 at 0:46

\begin{align*} &\text{For triangle ABC}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $r$ be the length of the inradius.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $p$ be the perimeter.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $k$ be the area.}\\[4pt] &\;\;\;{\small{\bullet}}\;\;\text{Let $h$ be the length of the altitude from B.}\\[12pt] &\text{Then}\;\;\frac{1}{2}pr = k =\frac{1}{2}|\text{AC}|h\\[4pt] &\implies\;pr = |\text{AC}|h\\[4pt] &\implies\;2(x+y+z)r = (x + z)h\\[4pt] &\implies\;\frac{r}{h} = \frac{x+z}{2(x+y+z)}\\[12pt] &\text{For triangle UBV, Let $h_1$ be the length of the altitude from B.}\\[12pt] &{\text{Then}\;\;h1}= h - |\text{XN}|\\[4pt] &{\phantom{\text{Then}\;\;h1}} = h - 2r\\[12pt] &\text{Then since triangle UBV is similar to triangle ABC,}\\[12pt] &{\phantom{\implies\;}}\frac{|\text{UV}|}{|\text{AC}|} = \frac{h_1}{h}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{h-2r}{h}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - 2\left(\frac{r}{h}\right)\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - 2\left( \frac{x+z}{2(x+y+z)}\right)\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = 1 - \frac{x+z}{x+y+z}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{(x+y+z) - (x+z)}{x+y+z}\\[4pt] &{\phantom{\implies\;}}{\phantom{\frac{|\text{UV}|}{|\text{AC}|}}} = \frac{y}{x+y+z}\\[4pt] \end{align*}

• where are you drawing the inradius from? and where did the equation $\cfrac{1}{2}pr$= $\cfrac{1}{2}$ $|{AC}|h$ come from? – Parley May 19 '17 at 20:21
• @Parley: The formula $$k = \frac{1}{2}pr$$ is a standard formula, and easily proved. And the formula $$k =\frac{1}{2}|\text{AC}|h$$ is just $$k =\frac{1}{2}bh$$ – quasi May 19 '17 at 20:27
• @Parley As I noted in my answer, to prove $k = pr/2$, consider the areas of triangle $IAB$, $IBC$, $ICA$ and their sum, $k$. Hope that helps! – fractal1729 May 19 '17 at 20:35