TITLE: Sum of angles under which a fixed line segment is seen from points situated on another line segment
QUESTION [10 upvotes]: I have a question, like a picture attached below. I can find each angle by sin( or cosine) rule, but I think there is an easy way...a clue ...a concept which made it easy. can someone help me?
I do appreciate any hint.

for example to find $A$ I use $$BC=\sqrt 2, AC=6 , AB=\sqrt {26}\\\cos(A)=\frac{c^2+b^2-a^2}{2bc}=\frac{36+26-2}{2*6*\sqrt{26}}$$ then find $A=11.3099$ and do like this for all the angles. But it is not the satisfying method. ( the gray squares are equal)
Thanks in advanced.

REPLY [18 votes]: 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)