Barycentric Coordinates of Incenter

Question

I'm trying to prove the fact that the incenter has the barycentric coordinates $(a,b,c)$. My reasoning goes like this; consider the triangle $\Delta ABC$ with barycentric coordinates $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ at $A$, $B$ and $C$, respectively. Let $|AB|=c$,$|AC|=b$ and $|BC|=a$. Let $D$ be the intersection between the internal angle bisector at $A$ and $BC$ and let $E$ the intersection between the internal angle bisector at $B$ and $AC$. According to the angle bisector theorem $\frac{BD}{CD}=\frac{c}{b}$ and $\frac{AE}{CE}=\frac{c}{a}$. If I'm correct that gives $D$ and $E$ the un-normalized barycentric coordinates $(0,b,c)$ and $(a,0,c)$, respectively. This is where I get stuck. How do I determine the intersection (ie the incenter of $AD$ and $BE$ from this. I am (as you can tell) very new to barycentric coordinates so please try to keep the answer simple. Thanks in advance!

Answer 1

Barycentric coordinates are a specific kind of homogeneous coordinates. In homogeneous coordinates you can join points and intersect lines both by computing the cross product. So you could compute $(A×D)×(B×E)$.

$$\left(\begin{pmatrix}1\\0\\0\end{pmatrix}\times \begin{pmatrix}0\\b\\c\end{pmatrix}\right)\times \left(\begin{pmatrix}0\\1\\0\end{pmatrix}\times \begin{pmatrix}a\\0\\c\end{pmatrix}\right)= \begin{pmatrix}0\\-c\\b\end{pmatrix}\times \begin{pmatrix}c\\0\\-a\end{pmatrix}= c\cdot\begin{pmatrix}a\\b\\c\end{pmatrix} $$

So if you have $c=0$, then the whole expression is the null vector, but otherwise it is just a representative of the coordinates $(a,b,c)$ as expected.

Answer 2

For triangle $ABC$,

• Let $a,b,c$ denote the lengths of sides $BC,CA,AB\;$, respectively.$\\[4pt]$
• Let $h_a,h_b,h_c$ denote the lengths of the altitudes from vertices $A,B,C$, respectively.$\\[4pt]$
• Let $r$ denote the inradius.$\\[4pt]$
• Let $k$ denote the area.$\\[4pt]$

Since the distance from the incenter to each of the lines $BC,CA,AB\;$is $r$, it follows that

• The "$A$" coordinate of the incenter is $\dfrac{r}{h_a}$.
• The "$B$" coordinate of the incenter is $\dfrac{r}{h_b}$.
• The "$C$" coordinate of the incenter is $\dfrac{r}{h_c}$.

From $$k = \frac{1}{2}\,a\,h_a = \frac{1}{2}\,b\,h_b = \frac{1}{2}\,c\,h_c$$ we get $$a = \frac{2k}{h_a},\;\;b = \frac{2k}{h_b},\;\;c = \frac{2k}{h_c}$$ hence $$ \frac{r}{h_a}:\frac{r}{h_b}:\frac{r}{h_c} = \frac{1}{h_a}:\frac{1}{h_b}:\frac{1}{h_c} = \frac{2k}{h_a}:\frac{2k}{h_b}:\frac{2k}{h_c} = a:b:c$$

It follows that the incenter has barycentric coordinates $(a,b,c)$, as was to be shown.

Answer 3

Let $[XYZ]$ be the area of triangle $XYZ$. Then by the areal definition, $[BCI]:[CAI]:[ABI] = \frac{BC \cdot r}{2} : \frac{CA \cdot r}{2} : \frac{AB \cdot r}{2} = a:b:c$ where $r$ is the in radius.