IMO Shortlist 1987 problem 21
Dodao/la:
arhiva2. travnja 2012. In an acute-angled triangle

the interior bisector of angle

meets

at

and meets the circumcircle of

again at

. From

perpendiculars are drawn to

and

, with feet

and

respectively. Prove that the quadrilateral

and the triangle

have equal areas.(IMO Problem 2)
Proposed by Soviet Union.
%V0
In an acute-angled triangle $ABC$ the interior bisector of angle $A$ meets $BC$ at $L$ and meets the circumcircle of $ABC$ again at $N$. From $L$ perpendiculars are drawn to $AB$ and $AC$, with feet $K$ and $M$ respectively. Prove that the quadrilateral $AKNM$ and the triangle $ABC$ have equal areas.(IMO Problem 2)
Proposed by Soviet Union.
Izvor: Međunarodna matematička olimpijada, shortlist 1987