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2 Madeline See, I think that was--the idea was that they were saying--Okay, I think we're over analyzing these things, but I think the idea was they were using the idea of having two points and finding a point that was equidistant from them in order to come up with if we look at the halfway point we can create perpendicular bisectors. "Oh look, the perpendicular bisectors meet at a point and oh, that point's equidistant from all of the vertices." So now that you have that knowledge, you were just given this knowledge of perpendicular bisectors, create a point that is equidistant from the vertices.
After that it can estimate the line AC as the chord of [D.sub.R]([A.sub.3]) and [A.sub.3] should lie on the perpendicular bisector of AC; that is, the location of [A.sub.3] should be (([x.sub.A] + [x.sub.C])/2, ([y.sub.A] + [y.sub.C])/2 - [square root of ([R.sup.2] - [([x.sub.A] - [x.sub.C]).sup.2]/4))] or (([x.sub.A] + [x.sub.C])/2, ([y.sub.A] + [y.sub.C])/2 + [square root of ([R.sup.2] - [([x.sub.A] - [x.sub.C]).sup.2]/4))].
and the perpendicular bisector. At this point they stated that triangle
Similarly, because the target position P is to the right of the perpendicular bisector of line AB, then we have:
In the sketch (Figure 3) [DELTA] ABC and [DELTA] FCB are reflections of one another in the perpendicular bisector of BC, which can be toggled on or off using the "Show perpendicular through H" and "Hide" buttons.
This exploration of the perpendicular bisector consists of two parts.
From here, students would again most likely resort to trial and error, but this activity motivates further discussion on perpendicular bisectors, thereby achieving a benefit from DG.
Choose [constructions] perpendicular bisector and click on the line segment AB.
It is worth noting that prior to or in the course of making this proof, students can fold perpendicular bisectors to the base in non isosceles triangles to "establish" that only in isosceles triangles does the perpendicular bisector pass through the opposite vertex.
Caption: Figure 9: Construction of the circle, using the perpendicular bisectors [P.sub.o][P.sub.v] and [P.sub.o][P.sub.f].