Thread: Bisecting a triangle with four chords of a circle

There are 4 chords in a circle. NK, NL, NM and KM.
C is the point in the center of the chord KM.
Also, the radius from the circle's origo O to point L is a line which passes through C and this line OL is perpendicular to chord KM.

Question:
Is chord NL always bisecting triangle KNM?
I.e., is angle KNL always = angle MNL?
Regardless of where N is on the circle? Given that all other points are fixed and it is "above" the chord KM as in the figure.


Methinks it should be, but I fail to derive it properly why it must. So help is asked!

Best regards

Yes, it's true. The Inscribed Angle Theorem states that for any chord, the inscribed angle is 1/2 of the central angle. For example, in your diagram, angle KNL is 1/2 angle KOL. In your case angle KNL = 1/2 angle KOL = 1/4 angle KOM.
Also angle MNL = 1/2 angle MOL = 1/4 angle KOM.

The Inscribed Angle Theorem is a good theorem to add to your arsenal. A good discussion is found at wikapedia. This theorem is in Euclid's Elements.