MathLabs

Problem 3

Let DD be an interior point of the acute triangle ABCABC with AB>ACAB>AC so that ∠DAB=∠CAD\angle DAB=\angle CAD. The point EE on the segment ACAC satisfies ∠ADE=∠BCD\angle ADE=\angle BCD, the point FF on the segment ABAB satisfies ∠FDA=∠DBC\angle FDA=\angle DBC, and the point XX on the line ACAC satisfies CX=BXCX=BX. Let O1O_1 and O2O_2 be the circumcenters of the triangles ADCADC and EXDEXD, respectively. Prove that the lines BCBC, EFEF, and O1O2O_1O_2 are concurrent.
Step 1 of 6: Show BCEFBCEF is cyclic via the isogonal conjugate of DD
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In plain words

The angle conditions defining EE and FF are exactly the conditions that make EE and FF lie on circles through DD and its isogonal conjugate.

AE⋅AC=AD⋅AD′=AF⋅ABAE\cdot AC=AD\cdot AD'=AF\cdot AB
Detailed analysis

Let D′D' be the isogonal conjugate of DD with respect to ∠BAC\angle BAC. The angle conditions ∠ADE=∠BCD\angle ADE=\angle BCD and ∠FDA=∠DBC\angle FDA=\angle DBC make quadrilaterals CEDD′CEDD' and BFDD′BFDD' cyclic, so a power-of-a-point computation at AA gives AE⋅AC=AD⋅AD′=AF⋅ABAE\cdot AC=AD\cdot AD'=AF\cdot AB; hence B,C,E,FB,C,E,F lie on a common circle.