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 1 of 5: Reduce to an isosceles-triangle statement

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In plain words

The line $AO$ is exactly the perpendicular bisector of the chord $FG$ of $\Omega$ .

AO\perp FG \ \Rightarrow\ (X\in AO \iff XF=XG)

Detailed analysis

Since $F,G\in\Omega$ and $AF=AG$ (both are radii of $\Gamma$ , which is centered at $A$ ), $A$ lies on the perpendicular bisector of chord $FG$ , and so does $O$ (as the circumcenter of $\Omega$ ). Hence line $AO$ is precisely the perpendicular bisector of $FG$ , and a point $X$ lies on $AO$ if and only if $XF=XG$ . It remains to show $XF=XG$ , equivalently $\angle KFG=\angle LGF$ where $X=FK\cap GL$ .