Circles $A,B,C$ are all of equal radius $r$ and are all tangent to each other. A smaller circle $D$ is tangent to all three circles on the inside (as shown above)
I'm well aware that triangle $ABC$ is an equilateral triangle, and I realize that angle $ADC$ must be $120°$ but this is what I would like to prove.
I tried proving this by extending line segment $AD$ to the line $BC$ and making the intersection point $E$ and attempted to prove the two triangles $AEC$ and $AEB$ are congruent, which in turn would mean $AD$ is a bisector of the 60° angle $CAB$ but I ended up with SSA (side-side-angle), and this is the one that fails to prove triangles are congruent. I also don't think we necessarily know yet that extending this line would create perpendicular lines (do we?)
Another thought was to draw an angle bisector in, but how do we necessarily know that this new segment would go through point $D$?
How can I prove that this angle is $120°$ ?
Hint:
Connect points $B$ and $D$ with a line segment and observe that $\triangle ADC$, $\triangle BDC$, and $\triangle ADB$ are congruent to each other.