 ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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# Constructing the Circumcenter - Problem 1

Brian McCall ###### Brian McCall

Univ. of Wisconsin
J.D. Univ. of Wisconsin Law school

Brian was a geometry teacher through the Teach for America program and started the geometry program at his school

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The circumcenter of a triangle is a point that is equidistant from all three vertices. The circumscribed circle is a circle whose center is the circumcenter and whose circumference passes through all three vertices.

In order to construct the circumscribed circle, first find the circumcenter of a given triangle. Find the perpendicular bisector of each side of the triangle. The circumcenter is the point of concurrency of the perpendicular bisectors.

Then, to draw the circle itself, place a compass at the point of concurrency and extend it to one of the vertices. This is the radius of the circumscribed circle. Then, swing an arc all the way around to draw the circumscribed circle.

Let;s take a look at an application of what we know about points of concurrency. In this problem it says find the point that is equidistant which means the same distance from AB and C.

So here I have points AB and C and I want to find the point that is equidistant from these three, but what are we going to do? Well we can start by asking ourselves if we found the circumcenter will that help? Well to find the circumcenter we’d have to do the three perpendicular bisectors, in order to do that we need a triangle. If I found that point of concurrency then it would be the center of this circumscribe circle, which would be equidistant from the three vertices.

So it sounds like if we found the circumcenter that will be the answer to our problem. So first thing I’m going to do is I’m going to draw in the three sides of this triangle. So I’m going to grab my straightedge and I’m going to connect A and C, I’m going to connect B and C and I’m going to connect A and B .

So if I want to find the point of concurrency of the three perpendicular bisectors we will have our circumcenter. So the first thing I’m going to do is I’m going to grab my compass and I’m going to bisect this side AC. So I’m going to swing an arc from point A making sure that my compass doesn’t move too much, I’m going to swing an arc from point C and I see that I have my perpendicular bisector. Now ideally I would have swung those arcs a little bit further apart but I can pick up where my two points are. So those two points are on my perpendicular bisector.

So I’m going to mark this as perpendicular and I’m going to mark these two segments as being congruent. We’re going to have to bisect one more side. So I’m going to choose side AB and I’m going to swing an arc from point A and I’m going to swing an arc from point B, and this is the trouble with doing these constructions. Sometimes it might get a little messy so you have to remember which point was which, so you’ve got that point and that point which will be on our perpendicular bisector.

Okay so I’m going to draw that line and I’m going to mark those two sides as being congruent and that being perpendicular and we could construct a perpendicular bisector of side BC but we know that this point would be concurrent of the three of them. So we found enough information to say that this point right here is equidistant from the three vertices. Now to prove that I’m going to draw in the circumscribed circle.

So I’m going to put the center of my compass and that point of concurrency and I’m going to extend it until I reach one of my vertices. So I know that this has to be the radius of this circumscribed circle. So I’m going to swing an arc all the way around to intersect B and I’m going to connect it to point A here and as you can see my compass slipped a little bit, or maybe I should use the present tense is slipping and we've constructed the circumscribed circle which is equidistant where the center is equidistant from three vertices and it passes through all three vertices.