proof that sum of interior angles in a plane triangle equals 180 degrees

Proof 1 (Euclid)
Let ABC be a triangle, and let one side of it BC be produced to D. Draw CE parallel to AB (Proposition I.31). Since AB||CE and AC has fallen upon them, the alternate angles BAC and ACE are equal (Proposition I.29). Also by Proposition I.29, since AB||CE and BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. It follows that the exterior angle ACD is equal to the sum of two interior and opposite angles (in triangle ABC) BAC and ABC:
ACD = CAB + ABC.

Add on both sides ACB. On the left we get two right angles; on the right the sum of the angles in DABC. Q.E.D.




Proof 2 (The Pythagoreans)

The Pythagorean proof is even simpler. A line parallel to the base BC is drawn through the vertex A, which gets us two pairs of alternate angles: CBA, DAB and BCA, EAC. Now, angle BAC complements, on the one hand, CBA + BCA to the sum of the angles of DABC, and, on the other, DAB + EAC to two right angles.