# Segment TrisectionWhat Is It About? A Mathematical Droodle

Explanation

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.

### Proof

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

(1)

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)

(2)

DK = DT / 3.

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

(3)

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.)

Copyright © 1996-2018 Alexander Bogomolny

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