Problem 4

Triangle

ABC

has circumcircle

\Omega

and circumcenter

O

. A circle

\Gamma

with center

A

intersects the segment

BC

at points

D

and

E

, such that

B

D

E

C

are all different and lie on line

BC

in this order. Let

F

and

G

be the points of intersection of

\Gamma

and

\Omega

, such that

A

F

B

C

G

lie on

\Omega

in this order. Let

K

be the second intersection point of the circumcircle of triangle

BDF

with segment

AB

, and let

L

be the second intersection point of the circumcircle of triangle

CGE

with segment

AC

. Suppose that the lines

FK

and

GL

are distinct and intersect at the point

X

. Prove that

X

lies on the line

AO

Step 3 of 5: Four parallelisms from directed-angle chasing

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In plain words

Repeated use of inscribed angles "anti-parallel" through the pair of lines $(FG,BC)$ produces four independent parallel pairs.

FD\parallel G_2C,\quad F_2B\parallel GE,\quad FB\parallel G_2E,\quad F_2D\parallel GC

Detailed analysis

Using directed angles modulo $180^\circ$ through the cyclic quadrilaterals $FBDF_2\subset(BDF)$ and $G_2ECG\subset(CGE)$ , together with $B,D,E,C$ collinear on line $BC$ : $\angle(FD,FG)=\angle(BC,GE)=\angle(G_2C,FG)$ gives $FD\parallel G_2C$ ; $\angle(F_2B,FG)=\angle(BC,FD)=\angle(GE,FG)$ gives $F_2B\parallel GE$ ; $\angle(FB,FG)=\angle(BC,GC)=\angle(G_2E,FG)$ gives $FB\parallel G_2E$ ; and $\angle(F_2D,FG)=\angle(BC,FB)=\angle(GC,FG)$ gives $F_2D\parallel GC$ .