I'm back again with another question regarding Cayley's Sextic.
Its geometric definition is as a pedal curve of a cardioid with the cusp being the pole. So, I draw the cardiod $r=2a\cos^2(\theta/2)$ with $O$ being the cusp, a tangent through a point $A$ and trace a perpendicular that passes through $O$, I get the point $P$ that draws the sextic.
Now, this curve has a trisection property: given any point on the cardioid, $\angle AOP$ is the half of $\angle AOD$. That's my question:
Given the geometric construction above, $\overleftrightarrow{OC}$ the bisector of $\angle DOA$, show that $\angle AOP \cong \angle AOC \cong \angle COD$
I've been trying the whole week on this problem, I've tried almost every triangle possible looking for a congruence but I always end up using something somewhat what I'm trying to prove. It doesn't seems like it's possible without using analytic geometry or some sort of calculus in polar coordinates.
Maybe I need something else in the given? A trick I'm missing? Striving for a contradiction instead? Any suggestion is greatly appreciated. Thank you all.
