 So recall that in congruence geometry, a circle is the set of all points, which have the fixed radial distance away from some point O based upon some radius O R. So basically we're looking for all points P on the circle, such that O P is congruent to O R. So we actually can define a circle without any notion of distance whatsoever. When we define circles in the previous video for lecture 21, we also prove that every line that intersects a circle can have at most two points of intersection. So let's take the maximum possibility. What if a line has two distinct points of intersection with the circle? Such a line is called a secant line to the circle because there's these two distinct points of intersection. Think of calculus one here with this term here. Now, when it comes to a secant, we could look at just the segment, just the segment of the line that lives inside the disc. Remember, you have the interior of the circle, those points which are less than the radius there, there's the circle itself, which are those which are congruent. The disc is when we take the circle and the interior together. If you take a secant line and intersect, if you intersect a secant line with the disc that it transverse, then you get what we call a cord or more specifically a cord of a circle is a segment whose endpoints are two distinct points on the circle. All right? A diameter of a circle is a cord that contains the center of the circle. So a diameter is gonna be something like this, okay? And so it's immediate to suggest that the segment sum of two radii is in fact the diameter of a circle, right? So if I take two different, if you take two different radii and put them together, you form a diameter. So I mean, after all, this right here is congruent to the radius OR so if you take your radius OR plus another radius OR, this is gonna equal a diameter of the circle, which we can also argue that a diameter is the maximum cord, but we'll talk about what that means in just a second. I'm getting a little out of myself here. A diameter is essentially two radii, two times OR. So this is the property that we've seen, of course, before that a diameter is two radii, okay? So that's what a segment sum of two radii and that's what a secant line is. What about a tangent line? Well, again, this is something we've probably seen before. So a tangent line would be a line that intersects the circle in exactly one point. Because after all, the intersection between a line and a circle is at most two points. So if it obtains the maximum intersections, a secant, if it only has one point of intersection, we call that a tangent line. And the point for which the single point is always called the point of tangency. Honestly, if you have a secant line, there's two points of intersection, you could call those the points of secancy. And this is actually where the word secancy came from when we talked about the secancy axiom for incidence geometry, that all lines have at least two points. We're thinking in this essence right here, a secant cuts a circle at two different points. You have to have at least two points in order for that to be possible, hence the word secancy from before, all right? And so with these notions of secant, core, diameter, tangent, we're ready to prove some important propositions with regard to circles in congruence geometry. So let's prove this first one here. If the segment AB is a diameter of the circle and CD is any other cord of that same circle, but is not a diameter, then it turns out that the segment AB is greater than the segment CD. And so this is what I meant when I said a moment ago that diameters are in fact, maximum, I should say maximal chords in that essence, no one is bigger than a diameter. But honestly, the word maximum would be appropriate right here because all diameters of the same circle are gonna be congruent to each other. All right, so how do we do this? All right, so we're gonna have a circle, we'll call it gamma, in which case I'll draw a very bad circle drawing over here of our gamma. And it's center, we're gonna call O. Usually you avoid using O as the name of the variable, but we like to do it for circles because we'd like to think of it as the origin, it's like the center of our geometry. So that's what O here stands for. And so consider the diameter AB. So the fact that it's a diameter means that the center of the circle is on the cord. So you have this cord AB, it contains O inside of it. And then you have some non-diametric cord, which we'll call CD for which let's say we'll put it over here or something like that. Here's C and here's D like so. So that's the setup. We have a diameter AB and we have a non-diametric cord CD. So then we're gonna apply the triangle inequality to the triangle COD, okay? So consider that triangle COD, which if we add that here, I like to spell things when I can. So this is triangle COD in case anyone likes name puns like that. But if you look at the triangle COD for a moment, by the triangle inequality, the segment CD is less than the segment OC plus OD. Again, that's just the triangle inequality in play here. But notice here that the segment OC and the segment OD are actually congruent to each other. And in fact, they're radii of the same circle. So OC is actually congruent to OA, which is a radius and OD is congruent to OB because it's also a radius. So OC plus OD is congruent to OA plus OB, which OA plus OB is in fact equal to the segment AB, all right? So I said that CD was non-diametric, but if we didn't assume that CD was non-diametric, if we just assume it was a cord, then with the triangle inequality, we do have to accept the possibility they have less than or equal to. But nonetheless, we show that every cord is less than or equal to a diameter and equality is only obtained when you have a diameter of the circle. So diameters are the largest cords of circles. Another proposition that we wanna prove with regard to circles is that if a diameter of a circle is perpendicular to a cord of the circle, then the diameter bisects the cord. So our picture is gonna look something like the following. We have a circle gamma right here. Let's say that the center of the circle is O. We have a diameter of the circle, which we're gonna call that one AB, just like we did before. So we have AB, we have, I'm sorry, we have AB right there. Then we're gonna take some non-diametric cord. So we got a picture like this. Here is C, here is D. And then by assumption, the diameter is perpendicular to the cord. So we have a right angle right here. In this situation, we then wanna prove that the diameter bisects the cord. Well, why are we assuming that it's non-diametric? Could it be a diameter itself? You could have something going on right here. Oh, sure, sure, sure. That's a possibility. But of course, that's sort of a trivial case in that situation because the center of the circle is going to be the midpoint of every single diameter that there is. And so if you have two diameters that intersect each other, like I have illustrated right here, then yeah, they're gonna bisect each other. That's no big deal whatsoever. I mean, you don't even need the right angle hypothesis in that situation. So that's why we're gonna fixate on a non-diametric cord because we don't necessarily know where the midpoint of such a segment is, okay? So the line AB is perpendicular to the line CD by assumption. And therefore the lines necessarily intersect. Let's call that point of intersection P, like we have here. So again, I want us to consider the triangle COD cod one more time. So the segments CO and OD are both congruent to each other because they're radii of the same circled gamma. Now again, we have this isosceles triangle for which then we know that the angle PCO is congruent to the angle PDO. And so then these triangles, you have right angles, of course, are congruent to each other, obviously. We know this already. So we have an angle angle side situation. So that these two triangles are congruent to each other. That was a consequence of the isosceles triangle theorem to get these angles to be congruent. But it turns out we didn't even need that as well. There's another congruent condition that's actually very appropriate to mention right here. The so-called hypotenuse leg congruence condition. This only applies to right triangles because in general, this is the side-side angle situation. Clearly, these two sides are congruent to each other and the segment PO is the same side. So it's congruent to each other and then right angles are congruent to each other, right? So if you have side-side angle, in general, that's not a triangle congruence. But for right triangles, if your side-side angle happens where, of course, the right angle is the angle that we know to be congruent, then one of the sides would have to be, since it's not angle, excuse me, since it's not side-angle side. For the right angle, the side-angle side would just be the two legs are congruent, which is perfectly fine. But with the right triangle, if it's not side-angle side, but it's side-side angle, that means one of the sides is the hypotenuse and then the other side is one of the two legs. So if we have a congruent hypotenuse and a congruent leg, then that actually means that the triangles are congruent to each other. And it's basically by the argument we just made right here. If you have these two triangles that satisfy the hypotenuse leg condition, you basically translate one of the triangles so that two of the legs glued together, the congruent legs, and so that way, the right angle are next to each other, so it forms a line. And then you play around with the assassin's triangle theorem like we just did and then argue these angles are the same and you end up with an angle-angle side situation. So that's basically the proof of the hypotenuse leg condition. So going forward, of course, whenever hypotenuse leg shows up, we can just use that as a triangle condition. In particular, as corresponding parts of congruent triangles are congruent, we then get that the segment Cp is congruent to the segment Dp, which then tells us that P is the midpoint of the segment CD and therefore the line AB, which is perpendicular to the segment CD, passes through the midpoint. Therefore, this diameter AB is a perpendicular bisector to the cord in that situation. So to end this video, I want to present two other propositions of circles and how diameters and cords interact with each other. And I'm going to just list them on the screen here and I'm gonna leave these as exercises to the viewer here because the proofs of these two propositions will be very similar to the propositions we've already considered in this video here. If we have a diameter of a circle that bisects a non-diametric cord of the circle, then the diameter is perpendicular to the cord. So this proposition is essentially just the converse of the one we just did here. We have a circle, we have a diameter, and we have some non-diametric cord such that the diameter bisects it. So these segments are congruent to each other because the diameter cuts it in half, like so. Well, then basically playing around with some triangle congruences inside of the circle, you're gonna argue this angle is a right angle because they're congruent to each other. So you have that one. And then the second proposition that I'm gonna mention but not prove is that the perpendicular bisector of a cord of a circle contains a diameter of the circle. So if we spell that one out here, we have our circle. This time we start with a cord of the circle, like so. The perpendicular bisector would look something like this and voila, that perpendicular bisector does contain a diameter. And so I'm gonna leave these as exercises to the student here.