If you are interested in practicing this technique on a few moderately difficult problems after reading this, go to the Geometer Problems Page.

Inscribe a circle in each of the three new triangles and connect the center of that circle to the opposite vertex of the original ΔABC. These three lines seem to meet at a point. Do they in fact meet there? If so, is it a known point? Can we prove it?

It's probably a good idea to draw this example yourself in
Geometer, but if you are lazy, you can download the file `Problem.T`. After drawing or
downloading the example, move the vertices A, B, and C and note
that the three lines from A, B, and C to the centers of the
opposite inscribed circles seem to meet at a point.

Even if you drew the figure yourself, to follow along it's
probably better to download the `Problem.T` file above so
that the labels, et cetera, will agree with the text below.

Here is the diagram you should see:

This is file: I:\html\geometer\Test\relations.txt Concurrent lines: (AA'') (CC'') (BB'') Concurrent lines: (BC) (AC) (l9) (CC'') Concurrent lines: (AB) (AC) (l7) (AA'') Concurrent lines: (AB) (BC) (l8) (BB'') Equal angles: (B A C) (A B v4) (A C v5) (C v6 B) (C v6 v4) (B v6 v5) (v4 v6 v5)...together with lots of other stuff. What's important is the fist line that indicates that, at least to the accuracy of the computer, the lines AA'', BB'' and CC'' seem to meet at a point in every configuration of the diagram that appeared while you moved the vertices A, B, and C. This is pretty good evidence that the lines are, in fact, concurrent.

**Note:** The way to scroll in the Relations window is to
click in it with the mouse and use the up-arrow and down-arrow
keys.

If you know anything about triangle centers (like the incenter,
circumcenter, orthocenter, centroid, et cetera), this point looks
like one. To see if it is one of the more common centers, you
can use the Geometer diagram `Finder.T` that comes with
the standard release.

If you don't have the standard release, download `Finder.T` here.

Load `Finder.T` into Geometer and you will see a triangle
with vertices A, B, and C. You need to construct your possible
new center on this triangle.

To save time, here is how I would do it:

- Set the "Line Type" to "Line"; this will make all your lines
go forever in both directions.
- Double-click on the "VL=>L Par" command. This will let you
repeatedly click on a vertex and a line and draw the line
parallel to that line through the vertex. Then click on A
followed by the segment BC, then B followed by AC, then C followed
by AB. Finally, click on the "Cancel Repeat Mode" button to get
out of repeat mode.
- Under the
**Primitives/Circle**pull-down, hold down the "Ctrl" key while you select the option**LLL=>C**. This will let you click on the three edges of a triangle to draw the circle tangent to all three (the incircle). Do this three times to make the three circles. Click on the "Cancel Repeat Mode" button to get out of this mode. - Under the
**Primitives/Vertex**pull-down, hold down the "Ctrl" key while you select the option**C=>V Ctr**. This generates the point at the center of a circle if you click on the circle. Do this three times to get the centers you want, and then cancel repeat mode. - Now connect the vertices of the triangle to the circle
centers appropriately, and find the intersection of two of those
lines. Change the color of that final vertex to "Cyan".
- Click on one of the lines you created and set the color to
"Invisible". Now click on the "Set Color" button and start
clicking on all the lines, circles, and points you drew except
for the final one that is in the color Cyan. Click on the
"Cancel Repeat Mode" button to get out of this mode.
- Finally, under the
**Edit**pull-down, select the option**Edit Geometry**. Go to the end of the file where the line for the final vertex should contain text that looks like:`[.cyan, .in]`. Edit this text to be:`[.cyan]`. Use the "Save File" (or Ctrl-S) option to save the file. At this point, your diagram should look just like it started, but with one new cyan-colored point that is the purported triangle center.

Now, to see if your point is a well-known triangle center, press the "N" key (or "Next" button). A bunch of triangle centers will appear. If one of them seems to match your point, wiggle the vertices A, B, and C, and check to see if it tracks your point. This should not happen the first time.

If there are no matches, press "N" again to look at another batch of centers. Continue in this way until you either exhaust the centers, or find a point that seems to match. In this case, you should find that your point is the same as the so-called "Speiker Center". If you look that up, you'll see that it is the incenter of the triangle formed by connecting the midpoints of the sides of the original triangle.

You may want to test this on the original figure in the diagram
`Problem.T`. In that figure, find the midpoints of the
triangle sides, form a triangle by connecting them, find the
inscribed circle, and finally, find its center. That point
should agree with the intersection of the lines. AA'', BB'', and
CC''. If you do this, quit the diagram without saving your work
and reload the original `Problem.T` if you wish to
continue the next section.

Knowing that your point is the Speiker Center may or may not help solve the problem, but at least it is interesting information.

We can again check, at least to the accuracy of the computer, by
drawing the line A''B'' and running the **Test
Diagram**-**End Test** commands as we did earlier. If we do
that, among lots of other information, we obtain these
interesting lines: (Remember that you can scroll in the
Relations window is to click in it with the mouse and use the
up-arrow and down-arrow keys.)

Parallel lines: (AB) (l5) (l9) (B''A'')

and

Equal length segments: [A B] [C v5] [C v6] [B'' A'']

This indicates that not only are the lines AB and A''B'' parallel, but they are the same length! That means that the quadrilateral AB''A''B is a parallelogram, and we know that the diagonals of a parallelogram bisect each other.

If we can show that, then we're basically done, since clearly BC will be parallel to and equal to B''C'', et cetera. Since we can form three different parallelograms using different pairs, each set of diagonals bisect each other, so all three must meet at the same point which is the midpoint of AA'', of BB'', and of CC''.

So how do we show that AB is parallel to A''B''? Well, A'' and B'' are the centers of circles tangent to the line through C parallel to AB, so they are the same distance from it (all three circles are of equal diameter, since all the three newly-created triangles are easily shown to be congruent). Thus A''B'' is parallel to a line known to be parallel to AB, so A''B'' is parallel to AB.

How do we show that AB = A''B''? Well, if we label the leftmost point in the diagram above X and the bottommost point Y, then since ΔXAC is congruent to ΔCBY, and B'' and A'' lie on the angle bisectors of all the angles of those triangles, we know that ∠XAB'' = ∠CBA'', so AB'' || A''B. Thus AB''A''B is a parallelogram and we are done.

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Email Tom: tomrdavis@earthlink.net.