 In this video, we're going to conclude our lecture 19 about midpoints and bisectors and we're going to prove the so-called best approximation theorem. So I want to remind you that in the previous video of lecture 19, we proved the so-called AOS relation, the angle opposite side relation, which tells us that if you have a triangle ABC, the angle A is going to be larger than the angle B if and only if you have that the side length BC is greater than the side length AC. So basically the bigger side coincides with the bigger angle and the bigger angle coincides with the bigger side. That's what we proved in the previous video here and we're going to use the AOS relation to prove a corollary of it known as this best approximation theorem. So best approximation actually says two things and it has to do with right triangles. So suppose that angle Q is a right angle for some triangle PQR. Then in that situation, we have that the angle P is less than Q and the angle R is less than Q as well. So in a congruent geometry, we have that for right triangles, the right angle is the biggest angle in the triangle. Now I want to put this in contrast to like elliptic geometry where you can have, for example, a triple right triangle. Best approximation would fail in such a situation because the right angle is not the biggest angle because they're all right angles. Why doesn't this not hold in elliptic geometry? Well, elliptic geometry is not a congruent geometry because it does have some issues with incidents and or betweenness. And in particular, in elliptic geometry, you don't have the alternate interior angle theorem, which from the alternate interior angle theorem, we get exterior angle theorems. We get the midpoint theorem, we get the bisector theorem, we get the angle opposite side relation. So all of these things are valid in congruent geometry, not in elliptic geometry because it's not a congruent geometry. But in a congruent geometry, it turns out the right angle is in fact the largest angle in the triangle. That's the first part. The second part of the theorem is going to say that if we have some point P, that's off of a line L. And if M is perpendicular to the line containing the point P, then it turns out that the intersection between L and M, call that point Q, if that's the point and then R is some other point on L other than Q, that's going to form your triangle PQR. Let me try to draw the picture here. So we have a line L, we have some line M that's perpendicular to it. Whoops, it's perpendicular to it. And so Q was the point of intersection. P was some point off of L. R is some other point over here. If we look at the line segment, the line segment PR is bigger than the line segment PQ. In other words, the closest distance from the point P to the line L is the perpendicular path. And so the second part is an immediate consequence of the first part because you have this right triangle. The angle Q is the biggest angle in this triangle. And so by the angle opposite side relation, the biggest angle coincides with the biggest side. So this side here is going to be bigger than the small side over here. So part B is just an immediate consequence of AOS using part A of the best approximation theorem. And this is actually where it gets its name, that if you're trying to find the best approximation, that is, what's the point on the line L that best approximates the point P? Well, it's going to be the perpendicular, the orthogonal projection. This has many important applications in geometry. For example, linear algebra use these things all the time. The orthogonal projection is the closest point to the point that's off of the linear space, or in fact it's off of the affine set like so. So the reason we use perpendicular distances, I should say the reason we use perpendicular lines to measure distance in affine geometry and Euclidean geometry is because it's the shortest distance. The least square solution in linear algebra has a consequence of this, which is why we call it the best approximation theorem. The best approximation to P on the line L is the point Q. And there's higher dimensional analogs of that, but we'll just stick with the two dimensional one right now. And so we've proven part B. So part B follows by part A. It follows from part A and the angle opposite side relation, which we've proven in the previous video. So let's then prove part A. So we have some right triangle. Oops, that's not a right angle, like so. I said that the right angle coincides with Q. So we'll put that label on this screen. And then P and R are just the other ones, just other points of the triangle. So we have this right triangle. We have to prove that the angle Q is larger than angle P and angle R. So let S be a point such that the angle PQS is an exterior angle to the triangle PQR. So we're just adding S to get an exterior angle right, for which then S would be over here somewhere. So here's S. So notice that S is an exterior angle. So the point PQS is we have this exterior angle right here. So by the exterior angle theorem, this angle SQP is going to be larger than the remote interior angles. So SQP is larger than angle P and angle R by the exterior angle theorem. But on the other hand, this is a right angle. So it's congruent to its supplement and therefore it has to also be a right angle from what we already saw here. So this tells us that angle Q is larger than the other ones as well. So this is a very simple theorem, very quick application of the exterior angle theorem and the angle opposite side relation that we've proven in this lecture and previous lectures and congruence. And that brings us to the end of lecture 19 about midpoints and bisectors. Thanks for watching this video. If you learned anything, please like these videos, subscribe to the channel to see more videos like this in the future, and please post any questions in the comments below if you have any, and I'll be glad to answer them. Bye, everyone.