Consider 3 collinear points $A$, $B$, and $C$, where $B$ is between $A$ and $C$. Draw two ellipses, one with foci at $A$ and $B$ going through $C$ and another with foci at $B$ and $C$ going through $A$. Suppose the two intersection points of the ellipses are $P$ and $Q$. Prove that $$\tan\left(\frac{1}{2}\angle PBC\right)=\frac{AB}{BC}$$
I tried letting the common focus, $B$, be the origin, and let $A=(-x_0, 0)$ and $B=(x_1, 0)$. Then the equations of the ellipses are:
$$\sqrt{(x+x_0)^2+y^2}+\sqrt{x^2+y^2}=2x_1+x_0$$ $$\sqrt{(x-x_1)^2+y^2}+\sqrt{x^2+y^2}=2x_0+x_1$$
I thought about subtracting and squaring, or using algebra to find the solution, but I am hoping there is a nicer solution, maybe using the focal reflective property of ellipses.
Apply the cosine rule to triangles $CBP$ and $ABP$, setting $\alpha=\angle PBC$, $a=AB$, $b=BC$, $x=BP$: $$ (2a+b-x)^2=b^2+x^2-2bx\cos\alpha\\ (a+2b-x)^2=a^2+x^2+2ax\cos\alpha\\ $$ Terms in $x^2$ cancel out and we are left with a linear system of equations in $x$ and $\cos\alpha$, which can be solved to yield: $$ \cos\alpha={b^2-a^2\over b^2+a^2}, $$ which is the same as $\displaystyle\tan{\alpha\over2}={a\over b}$.