PANAKIS' PSEUDOISOSCELES TRIANGLE

Let ABC be a trangle and D1, D2, D3 the feet of the internal angle bisectors [D1D2D3 = the cevian triangle of the incenter I]
Prove that the triangle D1D2D3 can be isosceles without ABC being isosceles.
Ioannis Panakis, Plane Trigonometry, vol. B, Athens (1973), p. 110 [in Greek]

I call this triangle ABC as Panakis pseudoisosceles triangle

Properties of ABC (in the same book pp. 109-111)
1. The A, D1, D2, D3 are concyclic [the cevian circle of I passes through A]
2. (a + b + c)*(-a^2 + b^2 + c^2) + abc = 0
3. a / (b + c) = b / (c + a) + c / (a + b)
4. (r1 - r) / (r1 + r) = ((r2 - r) / (r2 + r)) + ((r3 - r) / (r3 + r))
where r = the inradius and r1, r2, r3 the exradii.

PDF of the pages of the book Panakis

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