The curve in the picture is an ellipse with foci $A,B$ and center $C$. $P,Q$ lie on the ellipse. The line through $P$ is tangent to the ellipse and parallel to $VQ$ and $CD$. Thus, $CP$ and $CD$ are conjugate diameters. I have been told the following.
Apollonius's theorem
$$\frac{VP\cdot GV}{QV^2}=\frac{CP^2}{CD^2}.$$
I tried proving it and failed, I googled and found this completely unrelated theorem, I tried again and all I managed to do was to prove that the following implies my theorem:
$$CQ'\cdot CP'=CD^2.$$
To this end, note that $PQV\sim PP'C$ and $GQV\sim GQ'C$, because of sides lying on the same line or parallel lines. Hence:
$$\frac{VP}{QV}=\frac{CP}{CP'},\qquad\frac{GV}{QV}=\frac{CG}{CQ'}=\frac{CP}{CQ'},$$
hence, by substitution:
$$\frac{VP\cdot GV}{QV^2}=\frac{CP^2}{CD^2}\iff\frac{CP^2}{CP'\cdot CQ'}=\frac{CP^2}{CD^2}\iff CD^2=CP'\cdot CQ'.$$
And now I'm entirely stuck. I already tried describing this ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and thus parametrizing it as $\gamma(t)=(a\cos t,b\sin t)$, and finding all the relevant points, except the calculations got so horrible even Wolfram wouldn't compute, so I stopped. I tried constructing parallelograms to view both sides of these equations as areas or ratios of areas, but I can't see any way to go from there. So I'm completely stuck. How do I conclude?
Note
For the record, this is $CP'$, this should be $CQ'$, and luckily $CD^2$ is just $a^2\sin^2t+b^2\cos^2t$. Try multiplying those two roots…
Since @ArnaldoNascimento posted a link as a comment and the link answered the question, but did not convert it to a comment, I am doing so to get this off the unanswered list. However, I don't have time to copy the proof here, so I will sum-up the preliminary lemma, and put in an image of the proof of the desired result. If anybody has time, by all means I encourage them to copy it into the post.
The proof of this theorem resides at this link.
One first proves $CV\cdot CT=CP^2$, where $T$ is obtained by intersecting the tangent at $Q$ with the line containing $C,P,V$. Such a statement is easy in the case of a circle, where it is obtained by Euclid plus $CP=CQ$, and can be generalized to ellipses by dilating the ellipse figure into a circle figure, where the statement holds, and observing that proportions between segments on a single line are preserved.
Then, with a bunch of proportions and a couple extra points, one proves the desired result. Image: