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# Duplicating a Line Segment - Problem 2

###### 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

It is possible to create a triangle when given three line segments. First, duplicate the length of one line segment by using the compass to measure it and a straightedge to draw it. Then, measure another segment and, placing the compass at one of the endpoints of the triangle, swing the compass in an arc. Do the same with the other given line segment. Where these two arcs intersect is the third point of the triangle. So, by connecting the endpoints to this third point, you have drawn a triangle.

A common construction problem is given three line segments, create a triangle. So what you’re going to do is you’re going to duplicate these three line segments and then see where they intersect.

So we’ll start off by drawing a ray, so we have a base of our triangle. So I’m going to take my straight edge and I’m going to draw a ray. So we have one endpoint and we see that we’re being asked to construct triangle DEF.

So let’s start off by constructing our longest side which is side FD. So I’m going to take my compass and I’m going to duplicate this line segment FD onto my ray. So now that I’ve measured it, I’m going to make a mark on the end point and I’m going to come down and I’m going to duplicate that line segment down there. So I can label this endpoint as F and this end point as point D.

So we know that our third point E is somewhere up in space here. But how do we know exactly where it is? Well, we’re going to duplicate these segments, but we’re going to swing a big arc because we know that that arc has to be that length De. So we’re going to do that twice.

Let’s start off with DE. So sharp end on D I’m going to move my marker until it’s exactly on E. so I’ve duplicated that line segment. Now if the sharp end was here on D, I can’t go to point F, I’m going to have to go to point D. and I’m going to swing an arc from point D. so I know that this point D has, excuse me that point E is somewhere along that arc.

We’re going to do the same thing for this line segment EF. So I’m going to measure it again, and I’m going to put the sharp end on F and I’m going to swing my arc from F, and so where those two intersect has to be point D. So now you can mark this, excuse me I think I said point D I meant point E. but we’re not done yet because we don’t have a triangle. So I’m going to take my straight edge and I’m going to construct my last two sides. And so we have our second side of that triangle and finally we have our third side.

So the trick to this problem was realizing that if we constructed one segment and then swung the arcs from the endpoints, where they intersected had to be their third point.

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###### 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.

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