MathDB

Let

AB

be a diameter of a circle

(\omega)

with centre

O

. From an arbitrary point

M

AB

such that

MA < MB

we draw the circles

(\omega_1)

and

(\omega_2)

with diameters

AM

and

BM

respectively. Let

CD

be an exterior common tangent of

(\omega_1), (\omega_2)

such that

C

belongs to

(\omega_1)

and

D

belongs to

(\omega_2)

. The point

E

is diametrically opposite to

C

with respect to

(\omega_1)

and the tangent to

(\omega_1)

at the point

E

intersects

(\omega_2)

at the points

F, G

. If the line of the common chord of the circumcircles of the triangles

CED

and

CFG

intersects the circle

(\omega)

at the points

K, L

and the circle

(\omega_2)

at the point

N

(with

N

closer to

L

), then prove that

KC = NL

Problem Statement