Like the avatar by the way.

Diagram below.

Some preliminaries.

The centre O is on the intersection of the angle bisectors.

In triangles AFO and AEO, AO is common, OF = OE = radius and FAO = EAO so these are congruent. Therefore, AF = AE.

Similarly, BF = BD and CE = CD.

So triangles AEF, BFD and CDE are each isosceles.

Now to get an expression for angle EFD.

AFE = 90 - A/2 and BFD = 90 - B/2

=> EFD = 180 - (90 - A/2 + 90 - B/2) = (A + B)/2 = (180 - C)/2 = 90 - C/2

As C < 180 => C/2 < 90 => EFD is acute.

A similar argument can be used for the other two angles of DEF.

Bob

]]>I just have no idea how to do this. I think I should find the angles in terms of the larger triangles angles, but I duno how...:/

this thing is haaard. thanks!!!:D]]>