Every acute triangle is isosceles

Consider an arbitrary acute triangle ABC. Let E be the intersection of the bisector at vertex C and the bisection of the side AB. Let F and G be the projections of E on AC and BC respectively.

Since E belongs to the bisection of AB, we must have AE = BE. Also, since E belongs to the bisector of C, we must have EF = EG. However, this would imply that triangles AEF and BGF are identical, and then AF = BF. We also have that CF = CG, which implies that AC = BC. The arbitrary chosen triangle ABC is isosceles!

Can you find where the logic fails?

isoceles triangle puzzle

The bisector of C and the bisection of AB always intersect outside the triangle, on the circumcircle. One of the points F/G always lies on the segment AC/BC, and the other one does not.