How do I find the area of the inscribed triangle within the ellipse without analytic geometry? I have the following problem where I need to solve for the inradius of a particular inscribed triangle in an ellipse:

∆ABC is situated within an ellipse whose major axis is of length 10 and whose minor axis is of length 8. Point A is a focus of the ellipse, point B is an endpoint of the minor axis and point C is on the ellipse such that the other focus lies on BC. Compute the inradius of ∆ABC. Hint: recall the area formula for a triangle involving the inradius.

At the end I'm given a hint to use the area formula for a triangle $K = rs$  in solving for the inradius. I divided both sides by $s$ to get $K/s = r$, and then tried to solve for both $K$ and $s$. It was easy to solve for $s$, because if we let $A'$ be the other focus then
$$
\begin{equation}
\begin{split}
\begin{gathered}
P = AB + BC + AC = AB + BA' + A'C + CA\\
AB = a\\
BA' = a\\
A'C + CA = 2a\\
P/2 = 2a
\end{gathered}
\end{split}
\end{equation}
$$
However, I'm having trouble solving for $K$. I realize it's easy to just write up an analytic equation for the ellipse and find the equation of the line $BA'$ then solve for $C$, but I'm interested if there's a better way to do this. I noticed that since B connects line segments passing through both foci the question may have something to do with the reflective property of the ellipse, but I'm not sure. How do I find the area of the triangle non-analytically?

This isn’t the non-analytic solution you’re looking for, but the polar equation of the ellipse with a focus at the origin $$r={a(1-e^2)\over1-e\cos\theta}\tag{*}$$ provides a fairly easy way to compute the area of the triangle.  
Since $AC+BC=3a$, $BC=BA'+A'C$ and $BA'=a$, we can conclude that $AC=2a-A'C$, so we can use Heron’s formula to find the triangle’s area, which after simplification yields $K=a\sqrt{2A'C\cdot(a-A'C)}$.  
From (*) we can easily find that $r=a$ when $\cos\theta=e$, so $$A'C={a(1-e^2)\over1-e\cos(\pi+\theta)}={a(1-e^2)\over1+e\cos\theta}={1-e^2\over1+e^2}a.$$ Now, $e^2=1-b^2/a^2$, the semi-axis lengths are $a=5$ and $b=4$, and we have everything we need.  

Update: Taking an idea from Craig Hicks’ answer, we know that for $\angle B=2\phi$, $\cos\phi=b/a$ and $\sin\phi=f/a=e$, so $\cos2\phi=\cos^2\phi-\sin^2\phi=b^2/a^2-e^2=2b^2/a^2-1$. From $$AB^2+BC^2-2(AB)(BC)\cos(\angle B)=AC^2$$ we thus have $$a^2+(a+A'C)^2-2a(2b^2/a^2-1)(a+A'C)=(2a-A'C)^2$$ which simplifies to a linear equation in $A'C$. The solution to this equation is ${ab^2\over 2a^2-b^2}={ab^2\over a^2+f^2}$ which you can verify is equal to the previous expression for $A'C$ in terms of eccentricity.

Update 2: Blue’s comment above inspired me to find another expression for the triangle’s area besides Heron’s formula and $\frac12(AB)(BC)\sin(\angle B)$.  
Observe that the altitude from $A$ is the same for $\triangle{ABC}$ and $\triangle{ABA'}$, which means that $\operatorname{Area}(\triangle{ABC}):\operatorname{Area}(\triangle{ABA'})::BC:BA'$. The ”stretch factor” from $BA'$ to $BC$ is $1+{1-e^2\over1+e^2}={2\over1+e^2}$, so the big trangle’s area is ${2bf\over1+e^2}$.  
That’s about as geometric as I can make it.
