Given a triangle $ABC$ and angle bisectors $BD,CE$ which intersect at $O$ (incenter) . The ratio in which $O$ divides $BD$ is $3:2$ and it divides $CE$ in ratio $1:2$ . Find the ratio in which the third bisector is divided by the incenter .
[Math] Incenter divide ratio
geometrytriangles
Related Solutions
The trilinear coordinates of the incenter are $[1;1;1]$ and the trilinear coordinates of the $A$-excenter are $[-1;1;1]$, hence the barycentric coordinates of the $A$-excenter $I_A$ are $[-a;b;c]$ and
$$ I_A = \frac{-aA+bB+cC}{-a+b+c}=\frac{-|BC|(x_1,y_1)+|AC|(x_2,y_2)+|AB|(x_3,y_3)}{-|BC|+|AC|+|AB|}.$$
Two angles of $BAI_A$ are $\frac{A}{2},\frac{\pi+B}{2}$. Two angles of $BI_A P$ are $\frac{\pi-B}{2}$ and $\frac{A+B}{2}=\frac{\pi-C}{2}$.
It follows that in general $BAI_A$ and $BI_A P$ are not similar. This is not surprising: in your diagram, too, $BPI$ is acute-angled while $ABI$ is not.
Here is a possible picture for the already solved problem:
Well, how could you fill in the details without a faithful picture?! (And what is the "reflection of $N$ across $S$" by construction, the point $N$ is reflected with respect to the center $S$ or conversely?!)
Best Answer
In-center divides the angle bisectors in some definite ratio. Let $ABC$ be a triangle $AB = b$, $BC = a$ , $AC = c$ & let bisector of angle $A$ = $AF$ touching $BC$ at $F$ . Then $AF$ is divided in the ratio $(b+c): a$. Proceeding in this manner for other sides you will come to your answer . The answer will be $11:4$. Thank you.