I was playing in GeoGebra and came across a coincidence I could not explain:
A circle with center $C_1$ is internally tangent to a circle with center $C_2$ at point $S$, and externally tangent to a circle with center $C_3$ at point $R$. The point $C_3$ lies on the $C_2$ circle. In addition, circle $C_3$ intersects circle $C_2$ at P and Q. Extend R$C_1$ to meet circle $C_2$ at $U$. Next extend $RP$ and $US$ to meet at $T$. Prove that when points $S$, $R$, and $Q$ are collinear, then $\triangle RST$ is isosceles.
Please refer to the GeoGebra diagram if the problem statement was unclear. Link
We perform an angle chase. Let $\omega_1, \omega_2, \omega_3$ refer to the circles with center $C_1,C_2,C_3$ respectively. First, we start off with the following:
Claim. $C_1 C_3 Q C_2$ is a parallelogram (in that order).
Proof. Since, $S,R,Q$ are collinear, a homothety (i.e., dilation) at $S$ that sends $\omega_1$ to $\omega_2$ also sends $C_1$ to $C_2$ and $R$ to $Q$, so $C_1R \parallel C_2Q$. Furthermore, a negative homothety at $R$ sending $\omega_1$ to $\omega_3$ also sends $S$ to $Q$ (again because $S,R,Q$ are collinear), so $SC_1 \parallel C_3Q \Rightarrow C_1C_2 \parallel C_3Q$, which finishes.
(You can also phrase this part with only angle chasing, using the many isosceles triangles around.)
Now, angle chasing suffices. First, we have: $$ \angle TSR = 180 - \angle USQ = 180 - \angle UC_3Q = \angle C_2QC_3, $$ (where the last equality is by using the parallelogram proven above). We also have that $ \angle TRS = 180 - \angle PRQ $, but $\angle PRQ$ is equal to half the measure of the major arc $PQ$ (in circle $\omega_3$), so $180 - \angle PRQ$ is equal to half the measure of the minor arc $PQ$, or $\frac{\angle PC_3Q}{2}$. Now: $$ \angle TRS = 180 - \angle PRQ = \frac{\angle PC_3Q}{2} = \angle C_2C_3Q $$ so it suffices to show that $\triangle C_2C_3Q$ is isosceles which is immediate as $C_2C_3 = C_2Q$. (In fact, we have proven the stronger claim that $\triangle STR \sim \triangle QC_2C_3$!)