Find the perimeter of the triangle

Question

There is a triangle $ABC$. A line bisects the angle at the vertex $A$ and cuts the side $BC$ in point $D$ such that $BD=9$ and $DC=12$. If $O$ is the center of the circle that is inscribed inside the triangle $ABC$ and $AO:OD=4:3$, find the perimeter of the triangle $ABC$.

This seems very hard to me.

edit: I even drew one bad picture of this problem:

Answer 1

Since $AO$ is a bisector of $\Delta ADC$, we obtain:$$\frac{AC}{DC}=\frac{AO}{OD}$$ or $$\frac{AC}{12}=\frac{4}{3},$$ which gives $AC=16$.

Now, since $AD$ is a bisector of $\Delta ABC$, we obtain: $$\frac{AB}{AC}=\frac{BD}{DC}$$ or $$\frac{AB}{16}=\frac{9}{12},$$ which gives $AB=12$ and the perimeter is $$21+16+12=49.$$

Answer 2

I am not going to give the complete answer, but enough so that you can solve this yourself.

1. You already know BC (=BD+DC). So, you need to find AB and AC to find perimeter.

2. Angle bisector theorem should give you the ratio of AB:AC

3. The incenter divides AD in the ratio (AB+AC):BC.

(2) and (3) should give you the 2 equations for your to solve the problem.