Let the incircle of triangle ABC be tangent to sidesand at D, E, and F, respectively. Prove that triangle DEF is acute.
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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.
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