Suppose $e_1,e_2\in E$ and let $0\le t\le 1.$ There are $0\le \lambda\le 1;\ c_1,c_2\in C;\ d_1,d_2\in D$ such that $e_1=(1-\lambda) c_1=\lambda d_1$ and similarly for $e_2.$ We need to show that $x:=(1-t)e_1+te_2\in E.$
Now, it is easy to see that $(1-\lambda)C$ and $\lambda D$ are convex sets.
$x=(1-t)e_1+te_2=(1-t)(1-\lambda) c_1+t(1-\lambda) c_2\in (1-\lambda)C.$
$x=(1-t)e_1+te_2=(1-t)\lambda d_1+t\lambda d_2\in \lambda D.$
Thus, $x\in (1-\lambda)C\cap \lambda D$, so $E$ is convex.
As for the geometric interpretation of these sets, take two intersecting disks and see what happens when you multiply one by $\lambda$ and the other by $\lambda$. Then, try strips. Etc. In short, play with shapes!