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Let $a$ and $b$ be real numbers where $0 < a< b<180$. Let $A$, $B$, $D$ be points so $A$-$B$-$D$.

Part 1:
Prove there exists a triangle $ABC$ where measure of angle $CAB$ is $a$ and measure of angle $CBD$ is $b$.

How do I prove part 1 without making use of the Euclidean Parallel Postulate?

I know that d(A,B)+d(B,D)= d(A,D) due to definition of betweenness so it is on a straight line. To establish the side of AC, I could construct a ray AC by Angle Construction Postulate on a half plane on the same side of ray AB that makes angle $a$ and perhaps draw a line parallel to ray AC to show that angle is B is greater than angle A. I'm not sure how to show this is a triangle.
Should I show the overlap of the half planes using definition for interior in order to show that or prove that the triangle exists? Or is just the definition of the interior of any angle enough to show that the triangle exists?

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This sounds like exactly like the parallel postulate (in the form given by Euclid).

enter image description here

If angle $a$ is smaller than angle $b$, then the two lines meet somewhere if we keep following them upward.

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I believe it's impossible to do this without using some form of the parallel postulate, because it seems like the statement doesn't hold true in non-Euclidean geometry.enter image description here

$A$, $B$, and $D$ here are on a diameter of a circle representing the Poincaré model of the hyperbolic plane, with $B$ at the center; the two arcs through $A$ and $D$ are the (hyperbolic) lines through those points orthogonal to the line $ABD$. If we choose $b=\angle CBD = 90^{\circ}$ then $C$ must be on the (hyperbolic, but also Euclidean) line through $B$ orthogonal to the diameter; but now choosing $a$ just slightly less than $90^\circ$ means that we don't have 'enough space' to put $C$ anywhere on that line to form the correct angle $\angle CAB$.

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So about choosing a point like the meridian and a equator and then another longitudinal line in order to construct a hyperbolic triangle. Would this make any sense? – user31284 May 18 '12 at 3:58
@user31284. I'm not quite sure what you mean by that; can you explain yourself any better? – Steven Stadnicki May 18 '12 at 4:29
Sure. How about if I take a point on the north pole and then have a line go through that that is perpendicular to the equator and then is it possible to choose points of my triangle: one being a point on the equator and then another being the north pole and then the third one being another point also on the equator to show this weak exterior angle theorem holds? I'm using Geometer's Sketchpad to do this (hyperbolic plane). – user31284 May 18 '12 at 4:36

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