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Proof: perpendicular radius bisects chord

CCSS Math: HSG.C.A.2

Video transcript

In a previous video, we've already shown that if we have some circle centered at O right over here, and that if OD is a radius, and it's a radius that bisects chord AC-- so bisects means that it kind of splits it in two, that AB is equal to BC-- we've proven in a previous video that OD will be perpendicular to AC. So we've proven that we can assume that is perpendicular. And that video, if you want to look it up, and you might want to prove it yourself just out of interest, but the video-- if you do a search on Khan Academy for radius is perpendicular to a chord, you'll hopefully find the proof of that. What I want to do in this video is go the other way. If we know that OD is a radius and that it is perpendicular to chord AC, what I want to do in this video is prove that it's bisecting it. So we're not assuming that it's bisecting it in this video right over here. We're just assuming that it's perpendicular. So we're essentially going to go the other way. Here, we started with the fact that it bisected it, and we established that they were perpendicular. That was in the previous video. Now we're going to start with the assumption that they're perpendicular and then prove that they bisect. And just like we did in that previous proof, we'll set up some triangles here since we know a lot about triangles now. And we'll set up the triangles by drawing two more radii, radius OC and radius OA. And that's useful for us because we know that they're both radii for the same circle. So they have to be the same length. The radius doesn't change on a circle. So those two things are the same length. And you might already see where this is going. Let me label this point here. Let me call this M because we're hoping that ends up being the midpoint of AC. Triangle AMO is a right triangle. This is its hypotenuse. AO is its hypotenuse. Triangle OMC is a right triangle, and this is its hypotenuse right over there. And so already showed that the hypotenuses have the same length, and both of these right triangles share segment or side OM. So OM is clearly equal to itself. And in a previous video, not the same video where we explained this thing. In a previous video where we talk about-- and I think you can look it up. I think the video is called "More on why SSA is not a postulate." In that video, we say that SSA is not a postulate. So SSA, not a congruency postulate. But we did establish in that video that RSH is a congruency postulate. And RSH tells us that if we have a right triangle-- that's where the R comes from-- if we have a right triangle, and we have one of the sides are congruent, and the hypotenuse is congruent, then we have two congruent triangles. And if you look at this right over here, we have two right triangles. AMO is a right triangle. CMO is a right triangle. They have one leg that's congruent, right over here, MO, and then both of their hypotenuses are congruent to each other. So, by RSH, we know that triangle AMO is congruent to triangle CMO by the RSH postulate. And so if we know that they're congruent, then their corresponding sides have to be congruent. So based on that, we then know that AM is a corresponding side. Let me do that in a different color. AM is a corresponding side to MC. So we know that AM must be equal to MC because they're corresponding sides. These are corresponding sides. So congruency implies that these are equal. And if those are equal, then we know that OD is bisecting AC. So we've established what we need to do. Another way that we could have proven it without RSH, is just straight up with the Pythagorean theorem. We already know, just by setting up these two radii right over here, we know that OA-- so we draw a little line here. This is another way that we could have proven it. We already know that OA is equal to OC. And we also know that OM is equal to itself. OM is clearly equal to OM. It's equal to itself. And we also know from the Pythagorean theorem that AM-- do this in a new color. We know that AM squared plus OM squared is equal to OA squared. The length of the two legs squared summed up is equal to the length of the hypotenuse squared. So we know that for the left triangle right over here, for AMO. And we can set up the same relationship for CMO. We know in CMO that-- and I'm going to try to do it corresponding-- that CM squared plus OM squared is equal to OC squared. Now we know a few things. We know that OA is equal to OC. So, for example, right over here, where we have OA squared, we could replace this with OC. And then you can already see where this is going. You can already see that CM and AM are going to be the same. But if you want to do it a little bit more formally, you can subtract OM squared from both sides of this equation, and you'd get AM squared is equal to OC squared-- I've replaced this with OC squared-- minus OM squared. So that's on the left-hand side right over here. And then on the right-hand side, if we subtract OM squared from both sides, we have CM squared. CM squared is equal to OC squared minus OM squared. And then we could take the principal root of both sides of this because we really care about the positive root because we don't want to have negative distances. So if you take the principal root of both sides, this becomes AM is equal to the principal root of that. And we also get that CM is equal to the principal root of that. Well, these two quantities are the same. So AM must be equal to CM. They both equal to this quantity right here. So AM is equal to CM, and it might be a bisector. And this is really common sense. If you know that two sides of two different right triangles are going to be congruent to each other, you can always use the Pythagorean theorem to get the third side. And that third side is uniquely constrained by the length of the other two sides, because it's a right triangle. And so these are all ways of really getting at the same thing. But now we can feel pretty good about the fact that if OD is perpendicular here, then it's definitely going to bisect this chord.