My kid has this homework question:
Two circles whose centers are $M$ and $N$ meet at points $A$ and $B$. The point $N$ is on the circle whose center is $M$. The tangent at $A$ to the circle whose center is $M$ meets the circle whose center is $N$ at the point $C$ also. Prove that $AB=AC$.
I'm at a loss, as is my kid. We thought of proving angles $ABC$ and $ACB$ equal, or that $AB$ and $AC$ cut off equal arcs in their common circle, but don't see how to do either of those. I started playing with angles as described in the next paragraph, but it didn't seem to help any. How does one prove this? There's probably something very obvious that I'm just not seeing.
The aforementioned "next paragraph": Draw the tangent at $A$ to the circle centered at $N$, and say it meets the other circle at $D$ also. Then angles $ADB$ and $BAC$ are equal. Draw a line through point $A$ parallel to $BD$, meeting the circle whose center is $N$ at $E$. Then the angle that $AE$ makes with the tangent at $A$ equals angle $ADB$: but it also equals angle $ACE$, so $EC\Vert AB$.
By the alternate segment theorem, $\angle CAN=\angle ABN$. We also have $\angle ABN=\angle BAN$ and $\angle ACN=\angle CAN$ since $N$ is the circumcenter of $\triangle ABC$. So the isosceles triangles $\triangle ABN$ and $\triangle ACN$ are congruent and hence $AC=AB$.