About a property of the tangents to an ellipse from a given point

Question

I have recently stumbled upon this propriety of the tangents to an ellipse from a given point:

The tangents $PR$ and $PS$ from an external given point $P$ are equally inclined to the focal distances $PF_1$ and $PF_2$ (where $F_1$, $F_2$ are the foci of the ellipse). (see here, property 1).

I was wondering if the converse is also true:

Given four points $A,B,C,D$, such that points $C$,$D$ lie in the same semiplane determined by $AB$, let $I$ be the intersection of the external bisectors of angles $\angle ADB$ and $\angle ACB$. If $\angle CIA \equiv \angle DIB$, show that $C,D$ lie on an ellipse with foci $A,B$.

Multiple drawings in Geogebra suggest that this is true. However, i could not find a proof. Is this property real? If yes, how can we go about proving it?

Answer 1

Consider the ellipse with foci $A$, $B$ and passing through $C$. Line $CI$ is by construction a tangent of that ellipse and from any point $I$ of that line you can draw another tangent to the ellipse. By the property you cited, such a tangent must form with $BI$ an angle equal to $\angle AIC$: it follows that line $DI$, having that property by construction, is tangent to the ellipse.

And $D$ is the tangency point because it is the point on that line for which $AD+BD$ is minimum (that follows from $ID$ being the external bisectors of angle $\angle ADB$).

Answer 2

Let $A'$, $B'$ be points symmetric to $A$, $B$ with respect to $IC$, $ID$, respectively. Due to the assumptions on external bisectors, $A'$ lies on $BC$ and $B'$ lies on $AD$.

We have $A'I=AI$, $BI=B'I$, and $$\angle A'IB = \angle A'IC + \angle CIA + \angle AIB = 2\angle CIA + \angle AIB = 2\angle BID + \angle AIB = \angle AIB + \angle BID + \angle DIB' = \angle AIB'.$$

It follows that triangles $A'IB, AIB'$ are congruent (sas). Thus $A'B=AB'$, from which we get $$AC+CB = A'C+CB = A'B = AB' = AD+DB'=AD+DB.$$ Thus $C,D$ lie on an ellipse with foci $A,B$.