Draw the cone point down, with its vertex at the origin.
The points of the cone all have $0\le\theta\le 2\pi$.
Now that you've made that decision, you don't need to think about $\theta$ anymore. So from now on, just look at a general cross-section of the cone in which $\theta$ is constant. No matter what value of $\theta$ you choose, this cross-section is a right triangle in the $rz$-plane with height $h$ and base $a$.
Here's a picture: I've shaded the right triangle that I'm talking about. (Sorry: I don't have enough reputation to embed images.)
Image: triangular cross-section
The points of this triangular cross-section all have $0\le r\le a$.
Now you don't have to think about $r$ anymore, so from now on, just look at a cross-section of the triangle in which $r$ is constant. No matter what value of $r$ you choose, this is a vertical segment, as shown in the image above.
The points of that vertical segment all have $c\le z\le h$, and we need to find $c$. There are similar triangles in the figure, which you can use to show that $\frac cr=\frac ha$, and therefore $c=\frac ha r$.