The Angle Sum Problem: Solved

       I have found another method of proving the ∠ sum of △ that does not utilize ∠ properties of parallel lines!  This proof however, does not rely on the ∠ sum of △ being constant as my previous work on "The Angle Sum Problem" did, rather it proves that the ∠s of a △ are supplementary for a general case, thus proving it for all cases.  Consider the following △ (the line MM' intersects the perpendicular height and two non-base sides, through their midpoints).

Fig. 9 Scalene Triangle

        By separation of horizontal and vertical vectors, we can assume midpoints lie at the same height.

                  Let the Midpoint of AD be H

                  In △AHM' and △DHM'

                  AH = HD (by definition of midpoint)

                  HM' is common

                  AHM' = 90 (given)

                  DHM' = 180 - AHM' (supp ∠s on a straight line AD)

                             = 90

                 ∴ DHM' = AHM'

                 ∴ △AHM' ≡ △DHM' (SAS)

                 △AHM ≡ △DHM (similarly)

                 △AMM' ≡ △DMM'

                 ∴ ∠MDM' = ∠MAM' (corr ∠s in congruent △s)

                 ∠MDB = ∠MBD (base ∠s in isos △)

                 ∠M'DC = ∠M'CD (similarly)

                 ∠MDM' + ∠MDB + ∠M'DC = 180 (supp ∠s on straight line
                                                                               BC)

                 ∴ ∠BAC' + ∠ABD + ∠ACB = 180 (equal ∠s as proven above)

                 ∠ sum of △ = 180

         The basis of this proof is that, if you take each corner of any △ and fold them to the point where any interior perpendicular height meets its base; the angles will fit together on the base, a straight line that must form 180 degrees.