Ellipse problem: why are those two angles equal? I have been introduced, in class, to Newton's proof of the gravitational law. I was studying it, and one passage in the way it was presented did not convince me. So, I tried Wikisource's version of the Principia, where I found this picture, which is more or less the same picture I drew with GeoGebra by recomposing all the construction in the lesson notes:

The curve is an ellipse (the trajectory of the orbiting body), $P$ is where the body is, $A$ and $B$ are endpoints of the axes, $S,H$ are the foci, $C$ the center, $Qx$ is parallel to the tangent $PR$, $PF$ is orthogonal to $DK$, $DK$ is the diameter parallel to $PR$, $PG$ the one through $P$, $HI$ is parallel to the tangent line, $QT$ is orthogonal to $PS$, and $v$ is unclear to me. Referring to this figure, Wikisource's text states that $I\hat PR=H\hat PZ$. Why is that so?
 A: That's a basic property of ellipses: if you draw lines from each focus to a point on the ellipse, then they will make the same angle to its tangent at that point. Or, in optical terms: a light ray emitted from one of the foci and reflected in the mirrored inner surface of the ellipse will pass through the other focus.
The ellipse can be defined as the locus of all points that have the same total distance to the two foci.
If the tangent at $P$ did not make the same angle to $PS$ and $PH$, then moving $P$ a small distance towards, say $Z$, would increase the length $PS$ by more either more or less than it decreases $PH$, and in both cases $PS+PH$ would not stay constant.
A: Your question is equivalent to asking why is the angle SPR equal to HPZ? Recall that S and H are the foci and then consider SP and PH to be rays, with P the point of intersection.
I believe the equality of the angles SPR and HPZ then follows directly from the reflective property of an ellipse; that being, rays emanating from one focus will reflect from the boundary of an ellipse and then pass through the other focus.
The angles you mention are the angle of incidence and the angle of reflection of the ray, which are necessarily equal.
A: Look the ellipse like a generalisation of a circle (or a circle like a particular case of an ellipse). If those angles are not the same then when we put the foci together the tangent to the circle wouldn't be perpendicular to the radius.
