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Turkey NMO 2007 Problem 1, m(KNB)=m(BNL)

Source: Turkey NMO 2007 Problem 1

September 27, 2011

geometrycircumcirclegeometry unsolved

Problem Statement

In an acute triangle

ABC

, the circle with diameter

AC

intersects

AB

and

AC

at

K

and

L

different from

A

and

C

respectively. The circumcircle of

ABC

intersects the line

CK

at the point

F

different from

C

and the line

AL

at the point

D

different from

A

. A point

E

is choosen on the smaller arc of

AC

of the circumcircle of

ABC

. Let

N

be the intersection of the lines

BE

and

AC

. If

AF^{2}+BD^{2}+CE^{2}=AE^{2}+CD^{2}+BF^{2}

prove that

\angle KNB= \angle BNL

.

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