My uncle gave me the following puzzle, hoping there was a mathematical proof to his conjecture.
Let $A$, $B$, and $C$ be points on a circle of some fixed radius. The radius for $C$ bisects the chord $AB$.
Let $P$ be any point on the segment $AC$. Construct a second circle of the same radius which passes through $P$ and $C$. Let $Q$ be the point of intersection of this new circle and the line $BC$.
(See image below.)
My Uncle's Conjecture: $|AB| = |PQ|$.
I have created a proof which uses the Inscribed Angle Theorem a few times. The gist of it is below. Reference the following diagram.
Let $O$ be the center of the original circle, and let $\alpha = \angle COA$. Notice that also $\alpha = \angle COB$.
By the Inscribed Angle Theorem, we have $\angle CBA = \alpha/2$. By symmetry, $\angle CAB = \alpha/2$. Hence $\angle ACB = 180^\circ - \alpha$.
Next, let $R$ be the center of the constructed circle, and label $\beta = \angle CQP$ and $\gamma = \angle CPQ$. Then $\beta + \gamma = \alpha$.
Also, we have $\angle CRP = 2\beta$, and $\angle CRQ = 2\gamma$ by the same Inscribed Angle Theorem as before. Thus $\angle PRQ = 2\beta + 2\gamma = 2\alpha$.
Now, by SAS, $\Delta AOB \cong \Delta PRQ$. In particular, $|AB| = |PQ|$.
So, my question: Does this Theorem have a name or appear in the literature anywhere?