Draw A Triangle From 3 Excenters and Ex-radii

Question

My teacher gave me this problem and told me to think-

" Is it possible to draw a triangle, given the three ex-centers and length of the ex-radii?"

I don't know if it's possible or not. So, my question is-

1. Is it possible?
2. If yes, then how?

Also, he gave me another version where I'm only given the ex-centers, not the ex-radii

Answer 1

I think I found it

We are to find $\triangle PQR$. As @Blue said, $A,R,P$ are collinear. Similarly $A,Q,C$ and $C,P,B$ are collinear.

Note that, $RA$ is the external bisector of $\angle PRQ$ and $RH$ is the internal bisector of $\angle PRQ$. Hence they're perpendicular.

Now, $R$ and $H$ lies on the internal bisector. $C$ is an $excenter$, which is equidistant from the extension of $RQ$ and $RP$, which are the two sides of $\angle PRQ$. Hence $C$ lies on the internal bisector of $\angle PRQ$. So $R,H,C$ are collinear, and thus $CR$ is the height of $\triangle ABC$.

Thus follows my construction-

Given the excenters $A,B,C$, draw the triangle $ABC$. Drop perpendiculars from the vertices. The foot points of the perpendiculars are the 3 vertices of the required triangle.

Did I do it right?