# Trisect Segment: 2 Circles, 4 Lines

The diagram below illustrates trisections of a given segment $(AB)$ that requires drawing two circles and four lines:

### Proof

Observe that, by the construction, in the diagram below, $P$ is the midpoint of $CG$ while $E$ is the midpoint of $AG.$ It thus follows that $F$ is the centroid in $\Delta ACG,$ implying $AF:FP=2:1,$ but $P$ is also the midpoint of $AB.$ We therefore get the sequence of ratios: $AP:AF=3:2,$ $AB:AF=6:2=3:1.$

In the article referred to below a similar (but even shorter) argument was applied to $\Delta BCD,$ with $CE$ and $AB$ as the medians.

### References

|Contact| |Front page| |Contents| |Eye opener| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny