Results 1 – 24 of This worksheet is designed to replace a lecture on constructing the midsegments, perpendicular bisectors, and the circumcenter of a. Worksheet – constructing the circumcenter of a given triangle with compass and straightedge. Students will locate circumcenter of a triangle; Students will discover that the circumcenter is equidistant TI-Nspire calculator; Pencil; Straight-edge; Worksheet.
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So let’s apply those ideas to a triangle now. This distance right over here is equal to that distance right over there is equal to that distance over there. Let me draw it like this. And so you can construct this line so it is at a right angle with AB, and let me call this the point at which it intersects M.
Let’s prove that it has to sit on the perpendicular bisector. We have a hypotenuse that’s congruent to the other hypotenuse, so that means that our two triangles are congruent.
Circumcenter of a Triangle
So this distance is going to be equal to this distance, and it’s going to be perpendicular. What I want workwheet prove first in this video is that if workzheet pick an arbitrary point on this line that is workshdet perpendicular bisector of AB, then that arbitrary point will be an equal distant from A, workdheet that distance from that point to A will be the same as that distance from that point to B.
We’re kind of lifting an altitude in this case. And then we know that the CM is going to be equal to itself. Let me take vircumcenter midpoint, which if I just roughly draw it, it looks like it’s right over there. Then you have an angle in between that corresponds to this angle over here, angle AMC corresponds to angle BMC, and they’re both 90 degrees, so they’re congruent. So let’s say that’s a triangle of some kind.
And actually, we don’t even have to worry about that they’re right triangles. This length must be the same as this length right over there, and so we’ve proven what we want to prove.
So we can set up a line right over here. But if you rotated this around so that the triangle looked like this, so this was B, this is A, and that C was up here, you would really be dropping this altitude.
And the whole reason why we’re doing this is now we can do some interesting things with perpendicular bisectors and points that are equidistant from points and do them with triangles.
And I could have known that if I drew my C over here or here, I would have made the exact same argument, so any C that sits on this line. We know by the RSH postulate, we have a right angle. So if I draw the perpendicular bisector right over worksheeet, then this definitely lies on BC’s perpendicular bisector. This is going to be B.
So this side right over here is going to be congruent to that side. Just for fun, let’s call that point O.
And let me do the same thing for segment AC right over here. And now there’s some interesting properties of point O. This one might be a little bit better. Circumcenter of a right triangle. So let me just write it.
Circumcenter of a Triangle (with videos, worksheets, games & activities)
OC must be equal to OB. So to prove that C lies on the perpendicular bisector, we really have to show that CM is a segment on the perpendicular bisector, wokrsheet the way we’ve constructed it, it is already perpendicular. So this is C, and we’re going to start with the assumption that C is equidistant from A and B.
That’s what we proved in circjmcenter first little proof over here. So we’ve drawn a triangle here, and we’ve done this before. Although we’re really not dropping it.
So let’s call that arbitrary point C. So it must sit on the perpendicular bisector of BC.