Problem 3

Let

D

be an interior point of the acute triangle

ABC

with

AB>AC

so that

\angle DAB=\angle CAD

. The point

E

on the segment

AC

satisfies

\angle ADE=\angle BCD

, the point

F

on the segment

AB

satisfies

\angle FDA=\angle DBC

, and the point

X

on the line

AC

satisfies

CX=BX

. Let

O_1

and

O_2

be the circumcenters of the triangles

ADC

and

EXD

, respectively. Prove that the lines

BC

EF

, and

O_1O_2

are concurrent.

Step 4 of 6: Track the inverse circles

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In plain words

The inverse of each solution circle has a simple common chord with a fixed auxiliary circle.

A'=\operatorname{inv}_P(A),\qquad (ACD)\leftrightarrow(A'BD),\quad (EXD)\leftrightarrow(A'BX)

Detailed analysis

Let $A'$ be the inverse of $A$ . The inverse of $(ACD)$ is $(A'BD)$ because $B\leftrightarrow C$ , while the inverse of $(EXD)$ is $(A'BX)$ (using $CX=BX$ and the defining antiparallel relation). Let $Y$ be the radical center of $(ACD)$ , $(A'BD)$ , and $(ABC)$ ; their common chords show that $Y$ has the same power with respect to $(EXD)$ as well.