MathLabs

Problem 1

Let Γ\Gamma be the circumcircle of triangle ABCABC. Let DD be a point on side BCBC. The tangent to Γ\Gamma at AA intersects the line through DD parallel to BABA at point EE. The segment CECE intersects Γ\Gamma again at FF. Suppose BB, DD, FF, EE are concyclic. Prove that ACAC, BFBF, DEDE are concurrent.
Step 4 of 4: Radical axes are exactly the three lines
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radical axis(Γ,ω2)=BF, radical axis(Γ,ω1)=AC, radical axis(ω1,ω2)=DE\text{radical axis}(\Gamma,\omega_2)=BF,\ \text{radical axis}(\Gamma,\omega_1)=AC,\ \text{radical axis}(\omega_1,\omega_2)=DE
Detailed analysis

The radical axis of two intersecting circles is the line through their two common points. Hence the radical axis of Γ,ω2\Gamma,\omega_2 is line BFBF; of Γ,ω1\Gamma,\omega_1 is line ACAC; of ω1,ω2\omega_1,\omega_2 is line DEDE. By the radical center theorem, the three radical axes of any three circles are concurrent, so BFBF, ACAC, DEDE meet at one point.