Geometry puzzle with two nested triangles This "toy" problem actually came out of my research.
How does $\frac{AC}{BC}$ change as we vary $\theta$?
More specifically: let me use lowercase letters for distance from $C$. Let $\phi$ be the angle $A_1 C A_2$. I want to understand $\frac{a}{b}$ as a function of $\frac{a_1}{b_1}$ and $\frac{a_2}{b_2}$ as we vary $\theta$ from $0$ to $\phi$.
Is it possible to construct an example where
  $$ \frac{a}{b} ~ > ~ \alpha \frac{a_1}{b_1} + (1-\alpha) \frac{a_2}{b_2} \qquad \qquad (*) $$
where the point $A = \alpha A_1 + (1-\alpha)A_2$? In other words, $\alpha = \frac{c}{c+d}$ in the below picture.

Here's what I know so far: If $a_1 = a_2$ and $b_1 = b_2$, there is no example satisfying $(*)$, though it's not as obvious as it looks. (Proof: the right-angle triangles $CAA_1$ and $CBB_1$ are similar, so the ratio of $AC$ to $A_1C$ equals that of $BC$ to $B_1C$.)
Even if we assume $b_1 = b_2$ or $a_1 = a_2$, I don't know the answer yet.
(Two updates: some very rough simulations suggest there is no counterexample, and I think I may be able to sketch a very messy proof. Still looking for an elegant approach to this.)
 A: OK, I found the argument I wanted - eventually!
The answer is that $(*)$ never occurs: the ratio of the averages is smaller than the average ratio. Here is the proof.
Consider the ratio $f(d) = \frac{a}{b}$ as we scale $d$ up or down, rotating the ray through $A$ and $B$ to match. We actually don't need to compute $f$ (although trust me, I have and it's ugly). The important thing is that we write it as a function of $d$, not as a function of the angles.
The key observation is that, at least in the picture below, $f$ is a linear function of $d$. As we scale $d$, the lengths $a$ and $b$ are scaled linearly as well.

This picture covers case 1, where $\psi < \frac{\pi}{2}$. Here $f$ is completely linear in $d$.
The other case has a complication: at the point where the ray passes $90^\circ$, $f$ makes a change -- it becomes linear with a different rate of increase (or decrease).

To see what's happening, break the change in the ratio into a horizontal and vertical component. Nothing changes in the vertical direction. But before $90^\circ$, the horizontal change was "hurting" the ratio $\frac{a}{b}$ by decreasing both $a$ and $b$. After $90^\circ$ the horizontal change "helps" the ratio by increasing both $a$ and $b$.
What this argument shows is that after $90^\circ$, $f(d)$ is still linear but with a larger slope.

Above is a picture for the case where $f(d)$ is decreasing, i.e. as the $d$ increases the ratio $\frac{a}{b}$ goes down. But this doesn't change anything in the argument; the ratio is now going down more slowly.
..
The argument I just sketched (I'm not going to include the calculations) was that $f(d)$ is convex in $d$: It is either linear, or two joined-up linear functions with the second having a larger slope.
This fact proves my original claim as follows. $\frac{a_2}{b_2}$ is constant in this scenario. $\frac{a_1}{b_1} = f(d)$ for some $d$. And $\frac{a}{b} = f(\alpha d)$ for some $\alpha$. So
\begin{align}
 \frac{a}{b} &= f(\alpha d) \\
             &\leq \alpha f(d) + (1-\alpha) f(0) & \text{(convexity of $f$)}  \\
             &= \alpha \frac{a_1}{b_1} + (1-\alpha)\frac{a_2}{b_2}
\end{align}
which is what I wanted to show.
