# 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?

• Executing your construction in Cinderella, I don't see the claimed property. Maybe I misread what you wrote, but please double-check all your statements to reduce ambiguities, and add a figure of your own.
– MvG
Dec 16, 2017 at 22:45

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:

1. 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 $O_2$.

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).

3. 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 $\angle QAT = \angle PAM$).

4. 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 $\angle ATM + \angle ATS = \angle AQP + \angle AQS = 180^\circ$.

• I also noticed that the 2 points that the circles intersect each other at and the point at which MT intersects AQ are also collinear. How can I prove this?
– user482818
Dec 20, 2017 at 19:15
• @AndiQu Off the top of my head, instead of $MT$ think of $TS$ (we've already proven that it is the same line). $AQ$ is the perpendicular bisector of $TS$, so you want to prove that the midpoint of $TS$ lies on the segment connecting the two circle intersections; from then on, try again to show that the angle there is $180^\circ$. This might be doable with angles and arcs only, but I am not certain; I need to think on it. Dec 21, 2017 at 23:56
• @AndiQu, no, not with arcs; lengths are more useful here, I think. Try to prove that the foot of the perpendicular from the midpoint of $TS$ to $AB$ is the same distance from $A$ as the intersection of the common chord and $AB$ - that would mean they are the same point. The proof can proceed with similar triangles, or with trigonometry; in fact, the original question could also have been solved with trig. If you set this trig up in a Cartesian coordinate system, you would be taking what I call the "brute force" approach. Dec 22, 2017 at 11:50
• which similar triangles do you think I can use? All I got was that if $AT^2$ was equal to $AN$ times the distance from $A$ to the intersection of $AN$ and the common chord, then the statement is true
– user482818
Jan 2, 2018 at 7:09
• I am not seeing how you got the above, but if you are confident in that result, there may be a valid approach in it. Since $ATN$ is right-angled, $AT^2 = AN.AT_1$, where $TT_1$ is an altitude in that triangle. You may be able to prove that $T_1$ lies on the common chord. But I was suggesting something else: if $K$ is the midpoint of $TS$, and $KH \perp AB$, and $AC$ is a diameter, consider $AHK$ and $AQC$ to prove that $H$ lies on the common chord (that same property of right-angled triangles is needed here as well, applied to $ATQ$). Jan 2, 2018 at 12:52