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In the three dimensional figure below, is there a way to prove that $ \angle MNK = 90^ \circ $

$\hspace{2.8in}$diagram

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If I understand the labeling correctly, I don't think so—I don't see anything that prevents $\triangle MNL$ from "falling over" to land flat in the plane of $\triangle KNL$ (or being at any angle in between).

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There was a similar problem in a footnote by Heath in Euclid's Elements, I.8. Analogously, N was taken as the center of a circle with radius NM. Of course the angle of the 3d image, like your own, gives the circle the appearance of an oval. But if NM is perpendicular to NK, then MNK is 90 degrees.