How to geometrically prove the focal property of ellipse? How to prove geometrically that if we have a tangent of ellipse with focus F and F' in point P, that tangent is bisector of the angle between a line joining focus F to point P and the line F'P outside the ellipse?
 A: The answer of abel and Jantomedes above is the correct geometrical and easy construction, short of drawings. Here is a better presentation for it, on page 4, (in german), with a clear drawing, at http://www.mathematikunterricht.de/lehrplan/Planungen11.PDF.
I add the relevant page below, in case the original source disappears, and author is probably Monika Schwarze. 
There is also an easy analytical proof of the reflection property. I present it in short in case the original source disappears. Consider P, F1, F2 as vectors, . is scalar product. |x| is norm of vector.
We consider 
(1) |P-F1|+|P-F2|=const, which is a definition of ellipse. 
Assume P is parametrized as P(t), and differentiate by t. It does not matter, what t is actually. We receive, after not much calculation, with n as a vector in the direction of the tangent,
(2) n . (P-F1)/|P-F1| + n . (P-F2)/|P-F2| = 0.
This proves that cos(angles) are opposite, and relevant angles are equal.
This proof is from Omar Antolín Camarena : http://www.matem.unam.mx/~omar/notes/conicrefl.html

Below is the geometrical construction from Planungen11.PDF 

Translation of the German text
Special Tangent construction on $X$
Lengthen $F_1X$ by $F_2X$, find $G$, the axis $t$ of $F_2G$ is tangent to the ellipse through $X$.
Proof: all $Q\neq X$ and $Q\in t$ do not lie on the ellipse:


*

*$F_2Q=GQ$: $t$ is the axis;

*$F_1Q+F_2Q=F_1Q+GQ>F_1G=2a$.


Thus, $t$ halves the angle $F_2XG$ and the normal halves the angle $F_2XF_1$.
A: 
This expands upon and clarifies the correct answers above by Abel, Jantomedes, and Peter. Hopefully I have been very clear in my construction and argument.
The Reflective Property of an Ellipse:
A ray of light starting at one focus will bounce off the ellipse and go through the other focus.
Take ellipse $E$ with foci at $F_1$ and $F_2$. Take arbitrary point $P$ on $E$.
To prove: $F_1P$ and $F_2P$ make equal angles with the tangent line to $E$ at $P$.
Extend $F_1P$ past $P$ to $F_2'$ such that $F_2P=PF_2'$. Let $L$ bisect $\angle F_2'PF_2$ so that $\theta_1 =\theta_2$. $L$ is thus the perpendicular bisector of $F_2F_2'$, the mirror line for $F_2$ and $F_2'$.
Claim: $L$ is the tangent line to $E$ at $P$.
Suppose $L$ is not tangent to $E$ at $P$. Then $L$ meets $E$ at another point $Q$ on $E$. Since $Q$ is on $L$, $Q$ is equidistant from $F_2$ and $F_2'$. So $QF_2 = QF_2'$.
By the definition of ellipse, since $P$ and $Q$ are on $E$, $F_1Q + QF_2 = F_1P + PF_2$.
So $F_1Q + QF_2' = F_1Q + QF_2 = F_1P + PF_2 = F_1P + PF_2' = F_1F_2'$, which violates the triangle inequality in $\Delta F_1QF_2'$.
So $L$ is the tangent line to $E$ at $P$. $\theta_1=\theta_3$ by vertical angles, so
$\theta_3=\theta_2$, as desired.
A: here is my attempt. i wish i can post a figure to go with this but don't know how to make one.
let the two foci be $F_1$ and $F_2.$ just so you can visualize keep $F_1$ to the left of $F_2.$ 
(a) pick a point $P$ on the ellipse.
(b) extend $FP$ to $FP^\prime$ so that $F_2P = PP^\prime.$
(c) $M$ is the midpoint of $PP^\prime$ so that $PM$ ibisects the angle $\angle F_2PP^\prime$ and is the perpendicular bisector of $F_2P^\prime$  
(d) pick a point $Q$ between $P$ and $M$ on the line segment $PM.$
we will show that angle bisector $PM$ is the tangent to the ellipse defined by $F_1P + F_2P = k = constant$ by showing that $Q$ is outside the ellipse by establishing $$F_1Q = F_2Q > k \mbox{ for } Q \neq P \mbox{ on the line } PM.$$
proof of claim: $$k = F_1P + F_2P = F_1P + PP^\prime = F_1P^\prime< F_1Q + QP^\prime = F_1Q + F_2Q. $$
A: We're gonna prove the focal property by proving that a tangent bisects the given angle.
To do that, we're gonna bisect the angle with a line and prove that this line is the tangent by showing that there are no other points that belong to the ellipse on that line. It is pretty obvious that for ellipse in one point there can be only one tangent.
We do have the ellipse with foci $F_1$ and $F_2$ and a point $P$ on the ellipse curve.
Let's have the line $k$ which is gonna be the bisector of the $F_2PF_2'$ angle.
Let's extend a bit the ray $F_1P$ and place there a point $F_2'$ which is reflection of point $F_2$ in line $k$.
As $F_2'$ is the reflection, $PF_1+PF_2'=F_1F_2'=const=2a$ where $2a$ is the major axis.
If $k$ is the tangent, it intersects the ellipse in only one point which is $P$ but if it's not the tangent, there'll be a second point of intersection. Let's imagine a point $Q$ somewhere on the line $k$. Similarly, $QF_2 = QF_2'$ and since $Q$ is on the ellipse, $F_1Q + QF_2'=const=2a$.
Therefore, $F_1Q+QF_2' = F_1F_2'$ but from the triangle's possibility condition (well, I'm not sure what it's called in English) $F_1Q+QF_2' > F_1F_2'$ hence there cannot exist a point Q that belongs both to the ellipse and line $k$ other than $P$ so the line $k$ is the only possible tangent in point $P$.
That implies that the angle between the focus and the tangent $k$ is equal to the angle between $k$ and the second focus. That is the focal property of ellipse.
Little note 1: I forgot to hide point B on the image. It's useless.
Little note 2: The point $F_2'$ is called $F'$ on the image, my mistake again.

A: Suppose that you would like to find the shortest way from $A$ to $B$ that touches the angled line. Observe that if we were to reflect $B$, then the distance from $A$ to $B$ and from $A$ to $B'$ are the same, but the shortest path from $A$ to $B'$ is a straight line, the solution follows. The interesting part is that, because of the reflection, the angles marked red are equal.

Let $f(X) = |FX| + |F'X|$ so that the ellipse is exactly the set $f^{-1}\big(f(P)\big)$. Note that for any point $X$ outside of the ellipse we have $f(X) > f(P)$, on the ellipse we have equality, and inside $f(X) < f(P)$. Moreover, for any point $Y$ on the tangent setting $Y=P$ minimizes $f(Y)$, the sum of distances from $F$ and $F'$, exactly as in the first diagram. Hence, the marked angles are again equal.
I hope this helps $\ddot\smile$
A: It is known that a ray coming from F that is reflected in P will go throu F'. You will obviously get the same reflection if you replace the ellipes with it's tangent in point P. Therefore the normal of the tangent will bisect the angle of the incoming and outgoing ray. This implies that the tangent itself will bisect the supplement angle.
A: Distance $F_1 P + PF_2$ should be a constant in an ellipse.
In the language of calculus of variations the words constant and extremum have the same connatation or meaning.
Let foci be $ ( \pm c, 0) $ and let P move along a line.
Differentiate   $ \sqrt{ (x-a)^2 + y^2} + \sqrt{(x+a)^2 + y^2}  $ partially with respect to x and y to find out that $F_1 P + PF_2$  has equal inclination  to the line, which can later be recognized as its tangent to ellipse. (I am omitting the working details).
Better still,use an Optics method. Let $F_3 $ be reflection of $F_2$ about this line. The minimum time/distance of light travel is when $F_2 F_3 $ cuts the line, thus finding P's position. Vertically opposite angles and hence these incident angles which focal rays make to the tangent are equal.
This is a simplified situation of Fermat's principle. There are two media, sines of the angles is a constant (Refractive index). Special case here it is unity, the angles are same.
A: In the figure below one sees that the length of the path $FP_0F'$ is
equal to the major axis of the ellipse: $2a$ (ellipse definition).
If we take $P_1$ to be another point along the tangent line (in blue), the
length of the path $FP_1F'$ is longer than $2a$, since $\overline{FA}+\overline{AF'}=2a$ (ellipse definition) and $\overline{AF'}<\overline{AP_1}+\overline{P_1F'}$ by the triangle inequality.

In the figure below one sees that the path $FP_0F'$ has the same length
    as the straight line $FP_0F''$, where $F''$ is a reflection of $F'$.
    This path has minimal length, because, due to the triangle
    inequality, the distance $d(F, F'')\leq d(F,P_1)+d(P_1, F')$, in which $P_1$
    is any other point along the reflecting straight line.
    Hence, the minimal distance path implies that the incident, reflected
    and the transmitted ray angles are the same, as depicted in the figure.

