There is a rather involved (and it is claimed, well-known) proof that shows that all triangles are isosceles (it can be found in Euclidean and non-Euclidean geometries - Marvin Jay Greenberg, bottom of pg. 23) - but unfortunately after studying it, I cannot seem to find the flaw in the argument. Your help would be much appreciated. It claims:
Given triangle ABC. Construct the bisector of angle A and the perpendicular bisector of side BC opposite to angle A. Now consider the various cases (there are diagrams given in the book).
Case 1: The bisector of angle A and the perpendicular bisector of segment BC are either parallel or identical. In either case, the bisector of angle A is perpendicular to BC and hence, by definition, is an altitude. Therefore the triangle is isosceles (The conclusion follows from the Euclidean theorem that states: if an angle bisector and altitude from the same vertex of a triangle coincide, the triangle is isosceles.)
Suppose now that the bisector of angle A and the perpendicular bisector of the side opposite are not parallel and do not coincide. Then they intersect in exactly one point, D, and there are 3 cases to consider:
Case 2: The point D is inside the triangle Case 3: The point D is on the triangle Case 4: The point D is outside the triangle
For each case, construct DE perpendicular to AB and DF perpendicular to AC, and for cases 2 and 4 join D to B and D to C. In each case the following proof now holds:
(I don't have the appropriate symbol for congruence on my keyboard so I'll use '=' to mean congruence.)
DE = DF because all points on an angle bisector are equidistant from the sides of the angle
DA = DA, and angle DEA and angle DFA are right angles
Hence triangle ADE is congruent to triangle ADF by the hypotenuse-leg theorem of Euclidean Geometry. Therefore, we have AE = AF.
Now, DB = DC because all points on the perpendicular bisector of a segment are equidistant from the ends of the segment.
Also, DE = DF, and angle DEB and angle DFC are right angles.
Hence, triangle DEB is congruent to triangle DFC by the hypotenuse-leg theorem, and hence FC = BE.
It follows that AB = AC, in cases 2 and 3 by addition, and in case 4 by subtraction. The triangle is therefore isosceles.