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.
Solution
by Gerhard Wöginger
AKL and AML are congruent, so KM is perpendicular to AN and area AKNM = KM.AN/2.
AKLM is cyclic (2 opposite right angles), so angle AKM = angle ALM and hence KM/sin BAC = AM/sin AKM (sine rule) = AM/sin ALM = AL.
ABL and ANC are similar, so AB.AC = AN.AL. Hence area ABC = 1/2 AB.AC sin BAC = 1/2 AN.AL sin BAC = 1/2 AN.KM = area AKNM.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
2 Sep 1999