Proof - All Triangles are Isosceles

I shall now prove to you that all triangles are isosceles. That is, all triangles have two sides of the same length. No assumptions about the length of sides or size of angles are necessary for this proof to be clearly true. The word "scalene" can now be eradicated from our vocabulary as there are no scalene triangles. Here is the proof:

	Let ABC be any triangle.
	Draw the bisector of angle A as shown. Also draw the perpendicular bisector of the line BC as shown. The mid-point of BC is D. These two bisectors meet at M.
	Drop perpendiculars from M to AB and AC as shown.
	As AM is a bisector of A the lengths of the lines ME and MF must be equal. Call this length a.
	Draw the lines MB and MC.
	Because MD is a perpendicular bisector of BC the lines MB and MC must be of equal length. Call this length b.
	Consider triangle MFC. Using Pythagoras' Theorem, the length of FC must be (b²-a²). Call this length c. Similarly, using triangle MEB, the length of EB must be (b²-a²). This is length c.
	Consider triangle AEM. Let the length of AM = d. The length of AE = (d²-a²). Call this length e. Consider triangle AFM. The length of AF = (d²-a²). This is length e.
	Hence AB = e + c and AC = e + c So any triangle, ABC, is isosceles.
	In case you think it is because in some cases the point M falls outside the triangle it is quite simple to show the same methods will work. Again a line AM bisects angle A. DM is a perpendicular bisector to BC. Perpendiculars are dropped from M to AB and AC as before, but the lines need to be extended to meet these perpendiculars.
	As AM is a bisector of angle A, then ME and MF must be the same length. Call this length a. As DM is a perpendicular bisector of BC, then MB must equal MC. Call this length b. Consider triangle MFC. Using Pythagoras' Theorem, the length of FC must be (b²-a²). Call this length c. Similarly, using triangle MEB, the length of EB must be (b²-a²). This is length c.
	Consider triangle AEM. Let the length of AM = d. The length of AE = (d²-a²). Call this length e. Consider triangle AFM. The length of AF = (d²-a²). This is length e.
	Hence AB = e - c and AC = e - c So still, the triangle ABC is isosceles.

I have just conclusively proved all triangles are isosceles. Or have I?

Explain your thinking and how you go about finding the "error". If there is one. It may just be true!!!