Sum of angles under which a fixed line segment is seen from points situated on another line segment

The equal angles in the solution come from the fact that $BCDE, BCEF, BCFG, BCGH, BCHA, BIAJ$ are all parallelograms. This is because $BC,ED$ have the same slope and so do $BE, CD$, etc. Then, $\angle BDC=\angle DBE$ etc. by alternate angles. All the angles put together add up to rotating $BD$ onto $BJ$. Since $BJ$ is vertical and $BD$ is horizontal, they then add to $90$ degrees.

(Thanks @ACB for tidying up the image)

Hint 1:

The twelve line segments come in six parallel pairs.

Hint 2:

Move the angles around so that the parallel lines match up.

Solution:

Excuse my poor paint skills.