Prove that $|PF_{1}|+|PF_{2}|$ is Constant in an Elipse Given an elipse with two focus $F_{1}$ an $F_{2}$, and $A$ is an arbitrary point at the elipse. Stright line $AF_{1}$ has another intersection point $B$ with the elipse, and $AF_{2}$ has another intersection point $C$ with the elipse.
And point $P$ is the intersection of line $BF_{2}$ and $CF_{1}$, now the problem is to prove that $|PF_{1}|+|PF_{2}|$ is constant.
Thanks in advance!
 A: OK, imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$.  We draw an ellipse by holding a pen taut against the string.  The sum of the distances to a point on the ellipse from each of the tack points is
$$\sqrt{(x+c)^2+y^2} + \sqrt{(x-c)^2+y^2} = 2 a$$
The trick is to manage the algebra so that the derivation is readable.  First, square both sides to get
$$(x-c)^2 + (x+c)^2 + 2 y^2 + 2 \sqrt{x^2+y^2+c^2+2 c x} \sqrt{x^2+y^2+c^2-2 c x} = 4 a^2$$
This simplifies a little to
$$x^2+y^2+c^2+\sqrt{(x^2+y^2+c^2)^2-4 c^2 x^2} = 2 a^2$$
Now we need to rid ourselves of this remaining square root by isolating it:
$$\begin{align}(x^2+y^2+c^2)^2-4 c^2 x^2 &= [2 a^2 - (x^2+y^2+c^2)]^2\\ &= 4 a^4 - 4 a^2 (x^2+y^2+c^2) + (x^2+y^2+c^2)^2 \end{align}$$
We have some fortuitous cancellation which leaves us with a quadratic.  Rearrange to get
$$(a^2-c^2) x^2 + a^2 y^2 = a^2 (a^2-c^2)$$
or, in standard form:
$$\frac{x^2}{a^2} + \frac{y^2}{a^2-c^2} = 1$$
Note that, for an ellipse, $a>c$.  We interpret $a$ to be the semimajor axis, $c$ to be the focal length, and $b=\sqrt{a^2-c^2}$ is the semiminor axis.
To prove that the expression for the ellipse has the sum of the distances from the foci being constant, work backward from this sequence.
A: Let $AF_1=m,AF_2=n$,$BF_1=p,CF_2=q$.

Lemma$$\frac1m+\frac1p=\frac1n+\frac1q=\frac{2a}{b^2}$$

by def :$$m+n=p+BF_2=q+CF_1=2a$$
Using Menelaus' theorem and $CP=CF_1-F_1P$,we have
$$\frac{PF_1\cdot BA\cdot F_2C}{F_1B\cdot AF_2\cdot CP}=1$$
,we Omit tedious simplification
$$PF_1=\frac{\frac{4a^2n}{b^2}-2a-n}{\frac{4a^2}{b^2}-1}$$
Similarly,
$$PF_2=\frac{\frac{4a^2m}{b^2}-2a-m}{\frac{4a^2}{b^2}-1}$$
then
$$PF_1+PF_2=\frac{2a(4a^2-3b^2)}{4a^2-b^2}$$
