Trisecting an Angle Using the Cycloid of Ceva

Drag C.

∠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

.