I'm back again with another controversial puzzle for you.

If you try this puzzle for yourself, you will notice that it does not seem possible at first.

Please provide a proof for your solution. You will not need any Pythag or trig to solve this.


When a line from each side of a circle's diameter meet on the circle's circumference, they create a right-angle.

enter image description here

  • 1
    $\begingroup$ where is the puzzle? $\endgroup$
    – Oray
    Commented Jul 26, 2017 at 19:57
  • 3
    $\begingroup$ May I suggest, instead of a pretend context that is uninteresting, you introduce the mathematical concept about your puzzle that IS interesting? For example, is the intended solution not expected or does this appear impossible at first? $\endgroup$
    – Forklift
    Commented Jul 26, 2017 at 19:58
  • $\begingroup$ @Oray If you were able to find a solution which I missed, which did not provide a puzzling challenge, please post it and I may delete the post. $\endgroup$ Commented Jul 26, 2017 at 20:04
  • $\begingroup$ @Forklift That's a fair comment. I have changed the post accordingly. $\endgroup$ Commented Jul 26, 2017 at 20:06
  • $\begingroup$ Downvoted. Very easy even though not a puzzle. It should be on Math SE. $\endgroup$
    – newzad
    Commented Jul 26, 2017 at 20:37

1 Answer 1


Solution for the sake of completeness:

Angle x is 60 degrees. Opposite angles adding to 180 implies the quad is cyclic, which implies the bottom right angle below the red line is 30 degrees, which gives us our final answer easily.

The required mathematical knowledge is secondary school level, and proofs regarding the properties of, and how to identify

cyclic quadrilaterals

Can be found easily online or in textbooks.

  • $\begingroup$ Thanks for posting. I did not know about cyclic quadrilaterals. The challenge was to find the answer from scratch. I think if you had done it without knowing about them, you may have found it more puzzling. $\endgroup$ Commented Jul 26, 2017 at 21:10
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    $\begingroup$ Again this goes back to the statement "Just because something might be puzzling (to some) does not make it a puzzle" $\endgroup$
    – n_plum
    Commented Jul 26, 2017 at 21:16

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