Determine y-coordinate of a 3rd point from 2 given points and an x-coordinate. I'm working through the "Calculus 1" Coursera course (offline version, so no forums) and am struggling with the following question in the topic on Limits:

Consider points $A=(-10,-4)$ and $C=(8,5)$. The point $B$ is on the line
  passing through $A$ and $C$. The $x$-coordinate of $B$ is $-1$. Determine the
  $y$-coordinate of $B$. Hint: Do not figure out the equation of the line to
  solve this problem. Instead, use similar triangles to discover the
  equation of a line yourself.

The solution given starts by saying:

Draw your picture.
  $$\frac{\overline{BD}}{\overline{DA}}=\frac{\overline{CE}}{\overline{EA}}$$

The picture I drew was this (not a promising start).
They go on to say:

$\overline{DA} = x$-coordinate of $D$ minus $x$-coordinate of $A = -1 + 10= 9$
$\overline{CE} = y$-coordinate of $C$ minus $y$-coordinate of $E = |5+4|=9$
$\overline{EA} = x$-coordinate of $E$ minus $x$-coordinate of $A = 8 +
10=18$
...

I can see how we might derive a point $D$ as being some point with $-1$ as the $x$-coordinate, but I'm not following how they derive a point $E$.
I fear this is some basic geometry I've long-forgotten.
Thanks in advance!

The equation 
$$\frac{\overline{BD}}{\overline{DA}}=\frac{\overline{CE}}{\overline{EA}}\tag{1}$$
holds because $\triangle ACE$ is similar to $\triangle ABD$. 


I can see how we might derive a point D as being some point with -1 as the x-coordinate, but I'm not following how they derive a point E.

From the remaining information we conclude that the point $E(x_E,y_E)$ is the intersection of the horizontal line passing through $A$ with the vertical line passing through $C$, while the point $D(x_D,y_D) $ is the intersection of the horizontal line passing through $A$ with the vertical line passing through $B$ (see picture).
From $(1)$ and $\overline{DA} = 9$, $\overline{CE} =9$ and $\overline{EA} = 18$  we get  $\overline {BD}=9/2$. Since $\overline{BD}=y_B-y_D=y_B-y_A=y_B+4$, we find that the $y$-coordinate of $B$ is $y_B=1/2$.
