 In this video, we're gonna prove the following theorem of circles inside of congruent geometry. That is, we have the congruence axioms, the incidence axioms and the betweenness axioms. So this would also include neutral geometry, which you might have assumptions about continuity, but we won't need any continuity in order to prove this result. And this has to do with tangent lines with regard to circles. So what the theorem that we're gonna prove here is that a line is tangent to a circle, if and only if it's perpendicular to the radius drawn from the center of the circle to the point of tangency. So this is an if and only of statement. There's two directions that have to be gone here. So we have our circle, we're gonna call this gamma. And so what's called the center of the circle O and then the radius of this circle is gonna be something like OR, something like that. But we won't specifically mention that point right here. So we're gonna consider the situation. Which direction are we gonna go here? We're gonna assume that we have a tangent line and then we wanna argue that the intersection, excuse me, that the tangent line forms a right angle with the associated radius. So we might get something like the following. I'm gonna color my tangent line here in blue. So it just intersects the circle at one point. We're gonna call this line T for tangent line. And so then consider the segment that connects the center of the circle to this point of tangency. Let's call the point of tangency P and therefore OP is a radius of the circle. We then wanna prove that the line OP, we wanna prove the line OP is perpendicular to line T. So we're gonna do this by contradiction. We're gonna suppose not. Suppose that this line OP is not perpendicular to T. So the line segment isn't either. So then we're gonna let L be the unique line which contains O that is perpendicular to T. So we've proven previously that such a line does exist. So we're gonna take the perpendicular drop from the center of the circle. And again, for crudeness sake, I'm gonna exaggerate the situation here. So we're gonna call this line here L. But L forms a right angle with the line T and it does pass through the center of the circle. Let's call the point of intersection between T and L. We're gonna call that Q. This would then be the foot of the perpendicular that we drop from O onto T to use that vocabulary that we've used previously. All right? Well, by the extension axiom and also by the segment translation axiom, there has to exist a point R. Well, we're gonna let R be the unique point on T that's on the opposite ray of QP. All right? So QP is the ray that's going this way. So there's gonna have to be some point over here so that QR and QPR could grow into each other. So we have some point over here, R, so that the segment QR is congruent to QP, like so. So consider the angles PQO and RQO. These are both right angles and therefore they are congruent to each other. Those two angles are congruent to one another. So this then sets up for us a triangle congruence. For which triangle are we gonna consider here? So the idea is we have the triangle OPQ right here, but then we also have the triangle, which if I add in the missing side, we have the triangle OQR like so. And so these two triangles share a right angle, of course. They also share this side here, OQ, which is congruent to itself, and then by construction PQ and QR congruent to each other. So we have this side angle, side situation happening so that these two triangles are congruent to each other. And as corresponding parts of congruent triangles are congruent, we then can infer that the segment OR is congruent to the segment OP. Hold your horses there for a second. OP is a radius of the circle, okay? If that's a radius of the circle, that means that OR, it would have to also be a radius of the circle. In particular, that would distort our picture so that really R is this point right here, okay? In which case R is on the circle, P is on the circle, but that means this is in a tangent line because I have now two points of intersection. T was really a secant line, not a tangent line, and that gives us our contradiction. So therefore the line OP is perpendicular to line T. And so tangent lines are always perpendicular to their radius. I will then leave it as an exercise to the viewer to prove the other direction, which is very similar to what we just saw right here, that if a line intersects a circle and it's perpendicular to the radius associated to that point of intersection, then in fact that line was a tangent. And you'll prove this by contradictions because we have one point of intersection, it's either a tangent or a secant. Assume it's a secant and construct a contradiction, which will be similar to the picture we have on the screen right now. And so that brings us to the end of lecture 21 in our lecture series, which we were exploring congruence conditions on circles in a congruence geometry, just to get us used to circles. We've talked a lot about lines and angles and triangles and such, but circles are a very important part of geometry. So it was important that we then prove that many of the properties that we know about circles, in particular how circles intersect, how they interact with lines, behave in a congruence geometry like you would expect them to. And so these are not really Euclidean results, these are neutral geometric results. So thanks for watching. If you've learned anything about circles, please like this video, all the videos in this lecture in fact, subscribe to the channel to see more videos like this in the future. And if you have any questions, feel free to post them in the comments below and I will answer them as soon as I can. Bye everyone.