Solved Problem 3 - Expression for Inradius of the Incentral Triangle (of an Isoceles Triangle)

RESULT

For an isosceles triangle ΔABC with:
  •         sides a, b, and c, and b = c = ak, where k  is any positive real greater than 1/2
  •         area Δ
  •         internal angle bisectors AD, BE, and CF, and
  •         inradius r


Proof 
If X is the point of intersection of AD and FE (refer to above diagram):


XD = AD - AX


Applying the angle bisector theorem:
Using the above in XD, we get:
From the result for the side lengths of ΔDEF:
By the sine rule in ΔABC:
Returning to our expression for DE = DF and rewriting in terms of k:
FE is parallel to BC, so ∠EFC = ∠FCB = ∠ACF. Hence ΔEFC is isosceles.
We have already seen that:
We want an expression in terms of only r and k, so let us cancel out a by rewriting s as a(1+2k)/2:
Taking one (1+k) term inside the curly brackets:



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