Prove that 2 lines intersect on the circumference I have a geometry problem that asks to prove that 2 lines intersect on the circumference of a circle. It goes like this:
There are 2 circles, $O_1$ and $O_2$with centres $A$ and $B$ respectively, with the $A$ lying on the circumference of $O_2$. A point $P$ is chosen on $O_2$ so that it isn't in $O_1$. A line tangent to $O_1$ through $P$ meet $O_1$ at $S$, and it intersects $O_2$ again in $Q,$ with $Q$ and $P$ lying on the same side of $AB$. A line through $Q$ is tangent to $O_1$ again at $T$. A point $M$ is the foot of the perpendicular from $P$ to $AB$. Prove that $MT$ intersects $PS$ at $S$.
I tried to use tan-chord theorem and joining the intersections of $O_1$ and $O_2$ and joining $A$ and $B$ with the points of tangency, and then finding equal angles using isosceles triangles, but I couldn't get anywhere from there.
Is there a general way to prove that 2 lines intersect each other on the circumference?
 A: Basically, what you have to prove is that $M, T, S$ lie on the same line. Generally, co-linearity can be demonstrated in a number of ways, one of the simplest of which is with angles: show that $\angle ATM + \angle ATS = 180^\circ$.
STEPS:


*

*Let $QT$ intersect $O_2$ again at $N$. Prove that $N$ is the reflection of $P$ across $AB$ (alternatively that $\angle AMN = 90^\circ$, or that $\widehat{AP} = \widehat{AN}$; I'm using hats to denote arc measures). You can do this by considering the angles $\angle AQT = \angle AQS$ and the arcs they subtend on $O_2$.

*Show that $\Delta AQT \sim \Delta APM$ by showing that $\angle AQT = \angle APM$ (they subtend a common arc $\widehat{AN}$, which was the point of step 1). 

*Show that $\Delta AQP \sim \Delta ATM$ by using a side ratio from the similarity in the previous step, along with $\angle QAP = \angle TAM$ (which is just a rearrangement of $\angle QAT = \angle PAM$).

*Now you have $\angle AQP = \angle ATM$. Together with $\angle AQS = \angle ATS$ (which can be obtained in  many ways, such as for example showing that $ATQS$ is cyclic), you finally have $\angle ATM + \angle ATS = \angle AQP + \angle AQS = 180^\circ$.
