Triangle ABC is given. We extend side AC through C and mark point D on the extension. Angle DCB is by definition an exterior angle for triangle ABC. The angles of ABC to which it is not adjacent - angles A and B - are its remotes.

We will prove that the exterior angle DCB is greater than the remote angle B.

Mark the midpoint of BC; name it M. Construct the segment AN through M such that AM = MN. Connect N to C.

Angles BMA and NMC are vertical and thus congruent. M is the midpoint of segment BC and so BM = MC. AN was constructed so that AM = MN. Thus triangles BMA and CMN are congruent by SAS.

From this it follows that angles ABM and NCM are congruent. But angle DCB is the sum of angles DCN and NCM, and so angle DCB is greater than angle NCM. Thus angle DCB is greater than angle ABM, as was to be shown.

This technique can be used to prove that any exterior angle of any triangle is greater than its two remotes.

**Commentary**

The result proven holds in both Euclidean and hyperbolic geometry. But it does not hold in elliptic geometry. Where does the proof break down in elliptic geometry? We assumed that the segment from N to C is unique and that it lies in the interior of angle DCB. This cannot be assumed in elliptic geometry.

## No comments :

## Post a Comment