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. enter image description here

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.

  • 1
    $\begingroup$ As segments cut parallel lines, you either add them up at point $B$ or at point $C$. That I think gives you $90^\circ$. $\endgroup$
    – Math Lover
    Commented Oct 16, 2021 at 18:40
  • 4
    $\begingroup$ Where did this problem come from? (In particular, the image.) $\endgroup$
    – anon
    Commented Oct 16, 2021 at 18:41
  • $\begingroup$ Please, could you answer the question of runway44 ? $\endgroup$
    – Jean Marie
    Commented Oct 18, 2021 at 17:09
  • $\begingroup$ A question with a similar solution here. This question has motivated my proposal of a new title for your question. $\endgroup$
    – Jean Marie
    Commented Oct 18, 2021 at 17:16

2 Answers 2


Visual Solution

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)

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    $\begingroup$ @ACB Thanks so much! $\endgroup$ Commented Oct 17, 2021 at 15:41
  • $\begingroup$ Very fancy and lovely answer! $\endgroup$
    – MH.Lee
    Commented Oct 18, 2021 at 17:21
  • $\begingroup$ @Nightflight Thanks! $\endgroup$ Commented Oct 18, 2021 at 21:52
  • $\begingroup$ Fantastic mathematical insight! $\endgroup$ Commented Jun 22, 2022 at 17:29

Hint 1:

The twelve line segments come in six parallel pairs.

Hint 2:

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


enter image description hereExcuse my poor paint skills.

  • $\begingroup$ thank you, but can you clarify a bit more? please $\endgroup$
    – Khosrotash
    Commented Oct 16, 2021 at 18:35
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    $\begingroup$ @Khosrotash I've added the corresponding picture. $\endgroup$
    – Servaes
    Commented Oct 16, 2021 at 18:36
  • $\begingroup$ Thank you I got the idea. $\endgroup$
    – Khosrotash
    Commented Oct 17, 2021 at 4:54

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