Let's examine the centers of a triangle. We will start with an arbitrary triangle TriABC with it's sides extended (we'll see why later) in GSP.
Using GSP, we can simply select the sides of our triangle, and then select Construct Midpionts to obtain m1, m2, and m3 as shown.
Now we can select a vertex, say A, and the midpoint of the opposite
side, m2, and construct the segment between them. Do likewise for B and
m3, and C and m1. The three segments are concurrent and their intersection
is called the centroid of TriABC. We label this G by convention.
(Note: the GSP construction is not a proof.)
Now we can select our midpoints and our segments and "hide" them, in order to clean up our triangle before we move on. TriABC should now look like:
The next step is to select a vertex, say A, and the line opposite
A, which is the extended line through BC, and constructing a perpendicular
line from A through BC. Note that the intersection lies outside of
TriABC. Repeat the process for the other two vertices. The three
lines are concurrent and their intersection is called the orthocenter
of TriABC. We label this H by convention.
(Note: the GSP construction is not a proof.)
Now we can select our perpendiculars and "hide" them, in order to clean up our triangle before we move on. TriABC should now look like:
For our next step, first reconstruct the midpoints of our segments (just like
before). Then select a midpoint, say m1, and the segment the m1 lies on, in
this case AB. Once selected construct a perpendicular line using GSP.
This is called the perpendicular bisector of AB. Repeat the process
for the other two segments. The three perpendicular bisectors are concurrent
and their intersection is called the circumcenter of TriABC.
Here we label it C also.
(Note: the GSP construction is not a proof... just want this point to stick!)
Why is our newest point called the circumcenter? Because if we select the circumcenter to be the center of a circle whose radius is the length from the circumcenter to one of the vertices of TriABC, we obtain the circle that circumscribes TriABC, called the circumcircle. The picture below illustrates (it was cleaned first, can you duplicate it?).
The keen eye may notice that G, H, and C appear to be colinear. In fact,
they are. The following picture illustrates what is called the Euler line
segment for TriABC. Can you construct the line segment?
(Note: the GSP construction is not a proof)
Our final construction, is to construct the angle bisectors of each angle of
TriABC. The angle bisectors will be concurrent and their intersection is
called the incenter of TriABC, denoted I.
(Note: the GSP construction is still not a proof.)
The following pictures illustrates why I is called the incenter. It is the center of the circle that is inside TriABC and is tangent to each side. Can you construct the incircle?
The following illustrates the relationships between the triangle centers for various triangles.
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