Let AE, BF and CG be the angle bisectors of triangle ABC. Let the points E, F and G lie on BC, AC and AB, respectively. Let M denote the incenter of ABC.

Prove that if the sum of the areas of the inside triangles MCF, MAG and MBG equals half the area of triangle ABC, then ABC is isosceles. In otrher words:

area(MCF) + area(MAG) + area(MBG) = 1/2 * area (ABC)

=> ABC is isosceles. Will Ceva´s theorem and the angle bisector theoerem bring me any further?

## Prove triangle is isosceles

- Rainy Monday
**Posts:**20**Joined:**Sun Sep 25, 2011 3:55 pm

### Re: Prove triangle is isosceles

The incenter has three perpendiculars of equal length to each of the sides, so if the sum equals half the total area, AB + FC equals also half the perimeter of the triangle. The angle bisector theorem says AB/BC = AF/FC, meaning AB/BC = AF/(AF+BC-AB). Letting AB = X, BC = Y and AF = Z:

X/Y = Z/(Y+Z-X)

X/Y = [X + (Z-X)] / [Y + (Z-X)]

Either X=Y (which proves our theorem right) or Z=X (which proves AF = AB and FC = BC, which are impossible if ABC is to be triangle), so the first option is true.

X/Y = Z/(Y+Z-X)

X/Y = [X + (Z-X)] / [Y + (Z-X)]

Either X=Y (which proves our theorem right) or Z=X (which proves AF = AB and FC = BC, which are impossible if ABC is to be triangle), so the first option is true.