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Trisecting an Angle Using the Cycloid of Ceva

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∠ABD = ∠ABC/3
The cycloid of Ceva has the polar equation
r=1+2cos(2θ)
. To trisect the angle
ABC
, construct a line parallel to the polar axis (the positive
x
axis). Let
D
be the point of intersection of the cycloid and the line. Then the angle
ABD
is one-third of the angle
ABC
. Proof: let angle
ABD
be
θ
and let the point
F
on the
x
axis be such that
|BE|=|EF|=1
. Let
G
be the orthogonal projection of
F
on the line
BD
. The angle
DEF=2θ
, so
|BG|=1+cos(2θ)
. Since
|BD|=1+2cos(2θ)
,
|EG|=|GD|
,
|DF|=|EF|=1
. So angle
DFH
equals
4θ-∠EFA=3θ
, but
∠DFH=∠CBA
.
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