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Proof: radius is perpendicular to a chord it bisects

Sal proves that if a radius in a circle is drawn so it bisects a chord, then the radius is also perpendicular to that chord. The proof uses SSS congruence. Created by Sal Khan.

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Video transcript

In the last video, we learned that if we have two different triangles and if all of the corresponding sides of the two triangles have the same length, then by side-side-side, we know that the two triangles are congruent. And I also touched a little bit on the idea of an axiom or postulate. But I want to be clear. Sometimes you will hear this referred to as a side-side-side theorem, and sometimes you'll hear it as a side-side-side postulate or axiom. And I think it's worth differentiating what these mean. A postulate or an axiom is something that you just assume, you assume from the get-go, while a theorem is something you prove using more basic or using some postulates or axioms. Really in all of mathematics, you make some core assumptions. You call these the axioms or the postulates. And then using those, you try to prove theorems. So maybe using that one, I can prove some theorem over here. And maybe using that theorem, and then this axiom, I can prove another theorem over here. And then using both of those theorems, I can prove another theorem over here. I think you get the picture. This axiom might lead us to this theorem. These two might lead us to this theorem right over here. And we essentially try to build our knowledge or we build a mathematics around these core assumptions. In an introductory geometry class, we don't rigorously prove the side-side-side theorem. And that's why in a lot of geometry classes, you kind of just take it as a given, as a postulate or an axiom. And the whole reason why I'm doing this is one, just so you know the difference between the words theorem and postulate or axiom. And also so that you don't get confused. It is just a given, but in a lot of books-- and I've looked at several books-- they do refer to it as the side-side-side theorem, even though they never prove it rigorously. They do just assume it. So it really is more of a postulate or an axiom. Now with that out of the way, we're just going to assume going forward that we just know that this is true. We're going to take it as a given. I want to show you that we can already do something pretty useful with it. So let's say that we have a circle. And there's many useful things that we can already do with it. And this circle has a center, right here at A. And let's say that we have a chord in the circle that is not a diameter. So let me draw a chord here. So let me draw a chord in this circle. So it's kind of a segment of a secant line. And let's say that I have a line that bisects this chord from the center. And I guess I call it a radius because I'm going to go from the center to the edge of the circle right over there. So I'm going to the center to the circle itself. And when I say bisects it-- so these are all-- I'm just setting up the problem right now. When I say bisecting it, it means it splits that line segment in half. So what it tells is, is that the length of this segment right over here is going to be equivalent to the length of this segment right over there. I've set it up. I have a circle. This radius bisects this chord right over here. And the goal here is to prove that it bisects this chord at a right angle. Or another way to say it-- let me add some points here. Let's call this B. Let's call this C, And let's call this D. I want to prove that segment AB is perpendicular. It intersects it at a right angle. It is perpendicular to segment CD. And as you could imagine, I'm going to prove it pretty much using the side-side-side whatever you want to call it, side-side-side theorem, postulate, or axiom. So let's do it. Let's think about it this way. So you can imagine if I'm going to use this, I need to have some triangles. There's no triangles here right now. But I can construct triangles, and I can construct triangles based on things I know. For example, I can construct-- this has some radius. That's a radius right over here. The length of that is just going to be the radius of the circle. But I can also do it right over here. The length of AC is also going to be the radius of the circle. So we know that these two lines have the same length, which is the radius of the circle. Or we could say that AD is congruent to AC, or they have the exact same lengths. We know from the set-up in the problem that this segment is equal in length to this segment over here. Let me add a point here so I can refer to it. So if I call that point E, we know from the set-up in the problem, that CE is congruent to ED, or they have the same lengths. CE has the same length as ED. And we also know that both of these triangles, the one here on the left and the one here on the right, they both share the side EA. So EA is clearly equal to EA. So this is clearly equal to itself. It's the same side. The same side is being used for both triangles. The triangles are adjacent to each other. And so we see a situation where we have two different triangles that have corresponding sides being equal. This side is equivalent to this side right over here. This side is equal in length to that side over there. And then, obviously, AE is equivalent to itself. It's a side on both of them. It's the corresponding side on both of these triangles. And so by side-side-side, AEC. Let me write it over here. By side-side-side, we know that triangle AEC is congruent to triangle AED. But how does that help us? How does that help us knowing that we used our little theorem? But how does that actually help us here? Well, what's cool is once we know that two triangles are congruent-- so because they are congruent, from that, we can deduce that all the angles are the same. And in particular, we can deduce that this angle right over here, that the measure of angle CEA is equivalent to the measure of angle DEA. And the reason why that's useful is that we also see, just by looking at this, that they're supplementary to each other. They're adjacent angles. Their outer sides form a straight angle. So CEA is supplementary and equivalent to DEA. So, they're are also supplementary. So we also have the measure of angle CEA plus the measure of angle DEA is equal to 180 degrees. But they're equivalent to each other. So I could replace the measure of DEA with the measure of CEA. Or I could rewrite this as 2 times the measure of angle CEA is equal to 180 degrees. Or I could divide both sides by 2. And I say the measure of angle CEA is equal to 90 degrees, which is going to be the same as the measure of angle DEA because they're equivalent. So we know that this angle right over here is 90 degrees. So I can do it with that little box. And this angle right over here is 90 degrees. And because AB intersects where it intersects CD, we have a 90 degree angle here and there. And we could also prove this over there as well. They are perpendicular to each other.