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 4 of 5: The two quadrilaterals are homothetic

Previous step Next step

In plain words

Four pairs of parallel corresponding sides between two quadrilaterals force a homothety centered at the intersection of their supporting lines.

FBDF_2 \ \text{and}\ G_2ECG \ \text{are homothetic through } FG\cap BC

Detailed analysis

The four parallelisms of the previous step show that quadrilateral $FBDF_2$ and quadrilateral $G_2ECG$ have all four pairs of corresponding sides parallel ( $FD\parallel G_2C$ , $F_2D\parallel GC$ on one pair of sides, and $FB\parallel G_2E$ , $F_2B\parallel GE$ on the other), so they are homothetic, with center at the intersection point of lines $FG$ and $BC$ .