Let points $A, B$ and $C$ be on a circle $O$ such that segment $AB$ is a diameter of $O$ (meaning the center of the circle is the midpoint of the segment). Prove $m∠C = \frac{1}{2}$ (Internal angle sum(Δ$ABC$)).

Now this is not in Euclidean geometry and it is in neutral geometry. I don't know how to prove it and plus I'm limited since I haven't proven that the sum of the angles in a triangle add up to $180$ and the exterior angle inequality where the angle outside a triangle adds up to the two interior angles. Instead all I have is the Sacceri Legendre theorem (all angles of a triangle add up less than $180^{\circ}$ and exterior angles (angle on the outside is greater than either interior angles). I can use everything else.

Here is a picture:


  • $\begingroup$ "Neutral geometry" is simply Euclidean geometry but without the parallel postulate. Is this correct? $\endgroup$ – Fly by Night Feb 12 '13 at 21:44
  • $\begingroup$ No but its the same as absolute geometry. But im also limited since i can't use the the theorem that says angles in a triangle add up to 180 or the exterior angle theorem. $\endgroup$ – user60887 Feb 12 '13 at 21:46
  • $\begingroup$ According to Wikipedia "Neutral geometry" is another name for "absolute geometry". $\endgroup$ – Fly by Night Feb 12 '13 at 21:49
  • $\begingroup$ yes. Now how would i go about it? I tried drawing a tangent line through point c and a line from point c to the center R. But thats as far as i got. $\endgroup$ – user60887 Feb 12 '13 at 21:52
  • $\begingroup$ One minute "no", next minute "yes". $\endgroup$ – Fly by Night Feb 12 '13 at 21:53

Draw the line from the center of the circle to $C$. The two triangles are isoceles, so $m \angle C = m \angle A + m \angle B$

  • $\begingroup$ how would they be isosceles? Since C looks like lines up with point R that doesn't mean its isosceles. $\endgroup$ – user60887 Feb 12 '13 at 21:58
  • 2
    $\begingroup$ @user60887: $A, B,C$ are on the circle, so look at radii. $\endgroup$ – Ross Millikan Feb 12 '13 at 22:02
  • $\begingroup$ ok here is what i have so far i1091.photobucket.com/albums/i383/bonfire091/circle-1.jpg $\endgroup$ – user60887 Feb 12 '13 at 22:13
  • $\begingroup$ Now $\angle RAC =\angle RCA$ $\endgroup$ – Ross Millikan Feb 12 '13 at 22:17
  • $\begingroup$ I dont see it unless I can prove AC=BC then it doesn't fit the isoceles triangle theorem which states In triangle ABC if line segment AB is congruent to line segment AC iff angle B is congruent to angle C. $\endgroup$ – user60887 Feb 12 '13 at 22:24

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