From the purple triangle ABC and the center of its inscribed circle at D, form the three colored circles (BDC), (CDA), and (ADB), with centers F, G, H, respectively, thus forming the yellow triangle FGH. Prove that there is one circle that circumscribes both triangles.
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Now ∠AGC=2(α+γ)=180-2β=180-∠ABC. So ABCG is cyclic. Similarly we can show that AHBC and ABFC are cyclic, implying that AHBFCG is cyclic.