The case you describe is the general case since every measures $\mu$ and $\nu$ are absolutely continuous with respect to $\mu+\nu$. More precisely, there exists $h_{\mu,\nu}$ with $0\leqslant h_{\mu,\nu}\leqslant1$ everywhere such that $\mu=h_{\mu,\nu}(\mu+\nu)$ and $\nu=(1-h_{\mu,\nu})(\mu+\nu)$. Thus one can define an intrinsic product $\mu\odot\nu$ by
$$
\mu\odot\nu=h_{\mu,\nu}(1-h_{\mu,\nu})(\mu+\nu).
$$
When $\mu$ and $\nu$ are absolutely continuous with respect to the Lebesgue measure (or any other measure of reference) with densities $f$ and $g$ respectively, then $\mu\odot\nu$ is absolutely continuous with respect to the Lebesgue measure with density $f\odot g$ defined as follows: on $[f+g=0]$, $f\odot g=0$, and, on $[f+g\ne0]$,
$$
f\odot g=\frac{fg}{f+g}.
$$
This product $\odot$ on measures is commutative (good), associative (good?), the total mass of $\mu\odot\nu$ is at most $\frac14$ times the sum of the masses of $\mu$ and $\nu$, in particular the product of two probability measures is not a probability measure (not good?), $\mu\odot\mu=\frac12\mu$ for every $\mu$, and finally $\mu\odot\nu=0$ if and only $\mu$ and $\nu$ are mutually singular (good?) since $\mu\odot\nu$ is always absolutely continuous with respect to both $\mu$ and $\nu$.
Edit To normalize things, another idea is to consider $\mu\Diamond\nu=2(\mu\odot\nu)$. In terms of densities, this corresponds to a harmonic mean, since $\mu\Diamond\nu$ has density $f\Diamond g$, where
$$
\frac1{f\Diamond g}=\frac1{2(f\odot g)}=\frac12\left(\frac1f+\frac1g\right).
$$
In particular, this new intrinsic product $\Diamond$ is idempotent (good?), commutative (good), and not associative (not good?).
Edit A canonical product concerns probability measures and transition kernels. That is, one is given a measured space $(X,\mathcal X,\mu)$, a measurable space $(Y,\mathcal Y)$ and a function $\pi:X\times\mathcal Y\to[0,1]$ such that, for every $x$ in $X$, $\pi(x,\ )$ is a probability measure on $(Y,\mathcal Y)$. Then, under some regularity conditions, the product $\mu\times\pi$ is the unique measure on $(X\times Y,\mathcal X\otimes\mathcal Y)$ such that, for every $A$ in $\mathcal X$ and $B$ in $\mathcal Y$,
$$
(\mu\times \pi)(A\times B)=\int_A\mu(\mathrm dx)\pi(x,B).
$$
In particular, $B\mapsto(\mu\times\pi)(X\times B)$ is a probability measure on $(Y,\mathcal Y)$.
When $\mu$ has density $f$ with respect to a measure $\xi$ and each $\pi(x,\ )$ has density $g(x,\ )$ with respect to a measure $\eta$, $\mu\times\pi$ has density $(x,y)\mapsto f(x)g(x,y)$ with respect to the product measure $\xi\otimes\eta$.
You are asking for the interior angle $\angle CAE$ of an isosceles triangle $ACE$ with $AC = CE$ whose base $AE$ is on diameter $AB$ and opposite vertex $C$ on semicircle $AB$, such that another similar triangle $\triangle EDO \sim \triangle ACE$ with $D$ on semicircle $AB$ has $O$ at the midpoint of $AB$.
As such, let $\theta = \angle CAE$ and let $x = AE$. Assume without loss of generality that $AO = 1$, thus $EO = 1-x$. But $\triangle AOC \cong \triangle EDO$ because $\angle CAO = \angle CAE = \theta$ and $AO = CO = 1$, hence $\angle ACO = \angle CAO = \theta$. It follows that $AC = 1-x$. But by similarity, $$\frac{AC}{AE} = \frac{DE}{EO},$$ or $$\frac{1-x}{x} = \frac{1}{1-x},$$ hence $$x = \frac{3-\sqrt{5}}{2}$$ and $$\theta = \cos^{-1} \frac{1-x}{2} = \cos^{-1} \frac{\sqrt{5}-1}{4} \approx 1.2566370614359172954$$ radians.
Best Answer
A possible answer is given by S. Tabachnikov in his book "Geometry and billiards" and originally proposed by R. Peirone (Reflections can be trapped. Amer. Math. Monthly 101 (1994), 259–260).
First of all, one can construct a trap for light rays parallel to a given direction, see figure below (taken from Tabachnikov's book).
Curve $\gamma$ is an ellipse with foci $F_1$ and $F_2$, curve $\Gamma$ is a parabola with focus $F_2$. These curves are joined in a smooth way to produce a trap: light entering with a direction parallel to the axis of the parabola is reflected to focus $F_2$, and one can prove (exercise 4.3 in that book) that the path of reflected light through the foci of an ellipse converges to its major axis, so that a captured ray never escapes (provided the hole in the ellipse is not too large).
Unfortunately, such a trap only works if incident rays exactly have the given direction: it is shown in the same book (p. 115) that it is not possible to trap a set of divergent rays of light.
Edit by OP: the proof in the book that you can't capture a non-null amount of light is by Poincare Recurrence theorem; since the evolution of light-beams inside the container is given by a measure-preserving map. So Poincare Recurrence tells you that almost all ingoing light beams will eventually escape.
Edit. Just to clarify the above observation: some light CAN be trapped, but only if coming from parallel rays. Such light has vanishing volume IN PHASE SPACE (because all rays have the same momentum), hence that doesn't contradict Poincaré's theorem.