##### Like what you saw?

##### Create FREE Account and:

- Watch all FREE content in 21 subjects(388 videos for 23 hours)
- FREE advice on how to get better grades at school from an expert
- Attend and watch FREE live webinar on useful topics

# Constructing the Incenter - 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

The **incenter** is the point of concurrency of the angle bisectors. Since it is the center of the inscribed circle of a triangle, it is equidistant from all three sides. So, to find the incenter, first construct the angle bisectors of two of the angles of the triangle, and find their point of intersection. This is the incenter.

Using a compass, find the perpendicular distances between this point of intersection and each side of the triangle. This is the shortest distance between any point and a side, and is the radius of the circle. Using the compass, draw an arc of that size to construct the inscribed circle.

When you’re asked on a test to construct an incenter and the inscribed circle, first you have to ask yourself what is the point of concurrency that makes the incenter? Well let’s go back and take a look at the three keys of the incenter. It is the point of concurrency of the angle bisectors. It is the center of the inscribed circle, and therefore it’s equidistant from the three sides. So if we find the angle bisectors and see where they intersect, that will have to be the center of the inscribed circle. So let’s grab our compass and I am going to bisect one of these angles.

So I’m going to start off by swinging an arc from this angle on the lower right, so I’m going to swing an arc from there, have it intersect in two places and actually I don’t have a second point of intersection there. Okay there we go, and then I’m going from these two points of intersection, I’m going to swing two arcs and I can change my compass, you don’t have to I’m going to swing two more arcs so here is one arc, here is another arc and I’m going to draw in my line that forms my angle bisector. So I’m going to connect that point of intersection with the vertex and I have now bisected that angle.

So somewhere along this line is going to be my incenter. I’m going to bisect another angle here and actually I think it will be easier to bisect this angle right over here, so I’m going to start there I’m going to swing an arc and from each of these end points I’m going to swing two more arcs so I guess technically I’m swinging one arc from each end point, so I got my sharp end there, swing another arc and then now what I need to do is draw in my angle bisector connecting those two points.

Now here comes a great question Mr. McCall do you need to construct a third angle bisector? Well we know by our definition that’s the point of concurrency of the three angle bisectors. Point of concurrency means we’re going to have all three intersecting in the same spot. So in this triangle if three of them intersect here, then I know the third must intersect there as well, so I don’t need any more information. I have enough information to say that this must be the incenter.

The second part of this question says construct the inscribed circle, so what I’m going to do is I’m going to grab my compass and I’m going to have to find that perpendicular distance. So basically what I’m going to do if I were to sketch down below here right now I’m picturing a point and a line and I’m going to swing an arc and then from each of these two points, I’m going to swing two more arc, so I can find my perpendicular distance which is always the shortest distance between a point and a line.

So I’m going to grab my compass and I’m going to swing an arc that will intersect this bottom side of this triangle twice so swing an arc. I’ve got one intersection, I have my two points of intersection, so now I’m going to swing two more arcs I’m going to change this and make this a little bit smaller, I’m going to swing one arc right there and I’m going to swing another arc from that point of intersection and I’m going to draw in a perpendicular from the incenter to that side and I know that this will have to be the radius of my circle.

So the last step is to measure that distance and make it with my compass. Okay so I have my distance now I’m just going to swing around and create the inscribed circle. Okay well my compass is slipping a little bit so the circle is not going to be perfect, but on the test if you showed all of these markings, your teacher is going to show you full credit.

We bisected two angles. We didn’t have to do the third because we know that this has to be the intersection of all three. We found the perpendicular distance so I guess I should probably label that as a perpendicular and we said that has to be the radius of our inscribed circle.

Please enter your name.

Are you sure you want to delete this comment?

###### Brian McCall

B.S. in Chemical Engineering, University of Wisconsin

J.D. University of Wisconsin Law School (magna cum laude)

He doesn't beat around the bush. His straightforward teaching style is effective and his subtle midwestern accent is engaging. There's never a dull moment with him.

so my teacher can't explain this in 5 weeks but I learn this in less than 3 minutes”

its hard to focus when the teacher is really really really goodlooking”

i like how it took you 3 minutes and 8 seconds to accomplish what my teacher couldn't in 3 days”

##### Concept (1)

##### Sample Problems (1)

Need help with a problem?

Watch expert teachers solve similar problems.

## Comments (0)

Please Sign in or Sign up to add your comment.

## ·

Delete