Three tangents to a cardioid curve intersect at a point on its circumference

Question

While playing with GeoGebra a few days ago I came up with an initial property of my Cardioid curve, I heard this curve is well studied so I'm not sure on my guess if it's already discovered or new, adding references would be appreciated

You have a Cardioid and its circle of perpendicular tangents The red point is an arbitrary point of the circle of perpendicular tangents The violet line is a right angle bisector The blue line is the hypotenuse of a right triangle The green line is the reflection of violet with respect to blue The brown line is the reflection of blue with respect to green The brown line is tangent For giving me Cardioid.

I don't have progress on it yet, but I do have other cardioid guesses that I can include in a follow-up question on the site.

Answer 1

Your guess can be easily proved if we remember that a cardioid is also a caustic, tangent to the rays coming from a fixed point $P$ on a circle, after their reflection on the circle itself (see here for a geometric proof), in the sense that each ray is reflected about the line joining the centre of the circle with the point where the ray meets the circle.

A ray $PA$, for instance, is reflected to $AB$ (see figure below), which is tangent to the cardioid at point $Q$ such that $AQ/AB=1/3$. If $A'$ is the reflection of $A$ about the centre $O$ of the circle, ray $PA'$ is reflected to $A'B$, which is then also tangent to the cardioid. Hence from every point $B$ on the circle two perpendicular tangents to the cardioid can be issued, connecting $B$ to the endpoints of the diameter perpendicular to $PB$.

Consider now ray $PB$: by the above property its reflection $PD$ is tangent to the cardioid and $\angle PBO=\angle OBD = 90°-2\alpha$, where $\alpha=\angle ABP$. If $BE$ is the bisector of $\angle ABA'$ we have, on the other hand, $\angle EBP=45°-\alpha={1\over2}\angle PBO$. Hence $BE$ is also the bisector of $\angle PBO$ and $BD$ is thus the same as the brown line in your construction.