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 2 of 6: Use the tangent at DD
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In plain words

A tangent point creates equal powers for the two circles.

P=BC∩tD(BCD),PD is tangent to (DEF)P=BC\cap t_D(BCD),\qquad PD\text{ is tangent to }(DEF)
Detailed analysis

Let P=BC∩tD(BCD)P=BC\cap t_D(BCD). The defining angle conditions, together with DAB=CADDAB=CAD, give ∠PDE=∠DFE\angle PDE=\angle DFE; hence PDPD is tangent to (DEF)(DEF) at DD.