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### Course: Geometry (all content) > Unit 4

Lesson 4: Perpendicular bisectors# Circumcenter of a right triangle

Showing that the midpoint of the hypotenuse is the circumcenter. Created by Sal Khan.

## Want to join the conversation?

- how do you find the circumradius of a right triangle(5 votes)
- The smart-aleck answer: find the circumcenter of the triangle, then calculate the radius to one of the vertices.

The deeper answer: the circumcenter of a right triangle is the midpoint of the hypoteneuse. Thus the circumradius is half the length of the hypoteneuse.(9 votes)

- what does ratio mean in this video?(3 votes)
- The ratio used in this video was 1/2. I believe he use AB and BM.

1/2 over

BM/AB

In other words, if AB is equal to 2, then BM is equal to 1. If BM is equal to 5, AB is equal to 10.(7 votes)

- At what point did he prove OC was equal to OA and OB?(2 votes)
- Well, he proved this just about at4:05

Here is the logical argument: He has just proved that O is the midpoint of the hypotenuse, which means that OA = OB.

Then,

he reminds us of the initial construction of a perpendicular bisector to the leg BC.

A point on a perpendicular bisector is`equidistant`

from the endpoints of the bisected segment. (I just watched a video where he proved that.) Therefore, the distance from O to B will equal the distance from O to C.

In other words, that means that line segment OB = OC

Therefore`OA = OB = OC`

This is a reconstruction of the proof that OB would be congruent to OC:

Consider a line segment, PQ

construct a perpendicular bisector of PQ with line segment RS passing through point M, the midpoint of PQ.

Connect P and Q with R to make two back-to-back triangles. Now the original line segment PQ has been divided into two equal segments PM and MQ. These are congruent. A perpendicular bisector forms an angle of 90 degrees with the original line segment, so <RMQ = <RMP. The triangles we have constructed RMQ and RMP are congruent by SAS congruency postulate.

Finally, RQ must be congruent with RP by CPCTC

This is the proof that a point on the perpendicular bisector of a line segment would be equidistant from the points at the ends of the bisected line segment.(9 votes)

- what is cross multiplaction?(3 votes)
- Hi Deepti,

Cross multiplication is one of the ways to solve an equation when it is in the form of "a fraction equals a fraction". For example:

5/x = 22/3

We are trying to isolate x so one of the ways is to use our standard algebra methods. We have fractions so we want to get rid of those first. We'll get rid of the 3 by multiplying each side by 3:

(5/x)(3/1)=(22/3)(3/1)

15/x = 22

Now lets get rid of the x in the denominator:

(15/x)(x/1)=(22/1)(x/1)

15 = 22x

Finally, we divide by 22 to isolate the x:

15/22 = 22x/22

15/22 = x

The shortcut to all of this is to cross multiply. When we cross multiply, we multiply the numerator of one side with the denominator of the other side and vice versa:

5/x = 22/3

(5)(3) = (22)(x)

15 = 22x

This looks familiar :-)

So,

15/22 = x

Hope this help[s you to understand how cross multiplication works.

Good luck!(5 votes)

- Doesn't a bisector have to start from the center an angle? Sal drew it in the middle of the right triangle, not from the vertex from an angle. With the positions of angles in a right triangle, the only possible perpendicular bisector is the one protruding from the right angle.(3 votes)
- You are talking about the angle bisector.(2 votes)

- Couldn't you just draw it out? Wouldn't that be easier?(2 votes)
- It would, but drawings aren't exact, and you wouldn't know for sure that it bisects the hypotenuse(1 vote)

- How does he get the ratio and use it?(1 vote)
- OM is parallel to AC. Since both segments are perpendicular to BC, each triangle is a right triangle. Since they share an angle (B), and each have a 90 degree angle, the remainin angles, A and MOB must be congruent. Two triangles are similar if their corresponding angles are congruent. The corresponding sides of similar triangles are proportional.(2 votes)

- What does Circumcenter mean?(1 vote)
- The circumcenter is a two-part definition. Let's break it down, circum- means "circle". And center means "well a point that falls in the middle". So circumcenter is more like, the center of a circle. I know we are working on a triangle, but if you drew many circumradii from the circumcenter of the triangle, you are forming a circle on which the the triangle's vertices are touching.(2 votes)

- Cant you match up a circle to this one too? With the circumradius and everything...(1 vote)
- Yes. You can take the midpoint of the hypotenuse as the circumcenter of the circle and the radius measurement as half the measurement of the hypotenuse. Once you draw the circle, you will see that it touches the points A, B and C of the triangle.(2 votes)

- For what purpose, A perpendicular line is drawing over the triangle?(1 vote)

## Video transcript

What I want to do in this video is prove that the circumcenter of a right triangle, is actually the midpoint of the hypotenuse, and to do that, I'm gonna take, first take a look at the perpendicular bisector of one of the legs, of this, of this right triangle So, let me construct the perpendicular bisector of leg BC right over here, so it's going to look something like this, it's going to look something like this, this, it intersects at a right angle, its perpendicular, and it bisects it So B, this is from B to this point which we'll call M, maybe M for midpoint, is the same, as it is from M to C, so those two distances are going to be equal, and let's call the point where this perpendicular bisector intersects the hypotenuse, let's call this O,and we're gonna prove that O is the circumcenter of this right triangle Now, the first thing that you might realize, and this is what we've seen in many problems, the triangle OBM, looks similar to triangle ABC, and its actually not too hard to prove, they both already have a 90 degree angle, so if we show that they, they both have another angle Another set of corresponding angles that are congruent to each other, then we know that they're similar By AA similarity, and they both clearly share this angle, right over here, OBC is part of the smaller triangle, and ABC which is really the same angle, is part of the larger triangle, and so, and they also obviously share a 90 degree angle, so by AA triangle similarity, we have triangle OBM, OBM is similar, is similar to triangle ABC, is similar to triangle ABC, and what's useful about this? Is we know similar triangles are ratios between corresponding sides are constant, so for example, we know that the ratio between side BM, which is on the smaller triangle, we know that the ratio between BM Let me do this in a different color, just to, just for the sake of it, we know that the ratio between BM and BC, BM and BC, the ratio of this side on the smaller triangle to the corresponding side on the larger triangle Is going to be the same as the ratio of the hypotenuse on the smaller triangle, BO to the hypotenuse of the larger triangle, because they are similar, well we know what the ratio of BM to BC is, BM is half of BC, so this ratio over here is going to be equal to one half, this is M is the bi, of the midpoint of these things, so this is exactly the same distance as this, so this is one half of the entire BC, so if one half is equal to BM, over BC is equal to BO over BA We then know, if we just kind of ignore this middle part, right over here, that one half is equal to BO over BA, over BA, if you cross multiply it, if you cross multiply, you see that, well there's multiple ways to think about, but you could just cross multiply, and you say BA, is equal to 2BO, or if you divide both sides by two, and their really equivalent statements one half BA is equal to BO, so BO is one half of BA, so this is one half BA, and so this other length, AO right over here This is going to be B, this is going to be, this going to be BA, minus one half BA, so this is also going to be, one half BA, and so, this segment right over here, AO, AO is going to be congruent to OB So what we just shown, first of all, is that this perpendicular bisector, right over here, The perpendicular bisector of segment BC, it intersects the hypotenuse of our right triangle at the midpoint, So we've already established, so we, one thing that we've already established, is O, is the midpoint, is the midpoint, of the hypotenuse, of the hypotenuse, of the hypotenuse AB, well, that by itself is interesting, but, we also know that if a point sits on a perpendicular bisector of a segment, is equidistant, it's equal distant from the end point of the segment, we'd show that in a previous video So we also know that O, OB that's equidistant to the end points of the segment, right over here, that OB is equal to OC, but we know, from this first statement right over here, that OB is alsoequal to OA, OB is also equal to OA, its of OB is equal to OC OB is equal to OA, that means OC must be equal to OA, OC must be equal to OA Or another way to think about it, is at this point O, Is equal distant from all of the points on our tri, all of the vertices, I should say, this point O is equidistant, from all of the vertices of our triangle, of our triangle, So this distance, this distance, which is really going to become our circumradius, is the same as this distance right over here, which is the same as this distance right over there So that we know that O is equidistant, equidistant to all, all vertices, which is another way of saying that O is the circumcenter, O is the circumcenter, so we've just proven that if you have the circumcenter of a right triangle, it is the midpoint of the hypotenuse of the right triangle, or the other way around, that the hypotenuse of the right triangle is the circumcenter, because you only have one circumcenter of any, of any triangle