Segment Trisection
What Is It About?
A Mathematical Droodle

14 October 2015, Created with GeoGebra


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Copyright © 1996-2018 Alexander Bogomolny

Segment Trisection

The applet suggests the following statement:

From a point T on the base AB of ΔABC draw lines parallel to the medians AMa and BMb. Let D and H be the points of intersection of those lines with AC and BC, respectively. Then the medians AMa and BMb trisect the segment DH.

segment trisection


Let the notations be as in the diagram. In particular, G denotes the centroid of the triangle. This implies


GMb = BMb / 3.

Next note that similar triangles ABMb and ATD are divided by the median AMa into two pairs of similar triangles, from which with the help of (1)


DK = DT / 3.

Now, triangles DTH and DKE are also similar, so that (2) implies


DE = DH / 3.

Similarly, FH = DH/3. Therefore, DE = EF = FH.

(There is another examlpe of a serendipitous segment trisection that starts with a square inscribed into a quarter circle.)

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Copyright © 1996-2018 Alexander Bogomolny



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