THE 68th ROMANIAN MATHEMATICAL OLYMPIAD

Question

On the side [CD] of the square ABCD point E is chosen such that m( ABE) = 60^. On the halfline (BA point F is chosen such that [BE] = [BF]. Let M be the intersection between the lines EF and AD. a) Prove that m( BME) = 75^. b) The bisector of angle CBE intersects CD at N. Prove that the triangle BMN is equilateral. FIGURE_0

Accepted Answer

a) We have m( MEB) = m( CEB) = 60^, hence EB is an exterior angle bisector in the triangle MDE. But m( MDB) = m( BDE) = 45^, hence DB is an interior angle bisector in the triangle MDE. Therefore, MB is the bisector of AME and since m( AME) = 150^, it follows that m( BME) = 75^. b) We have BAM BCN, hence BM = BN. Also, m( BMN) = 90^ - m( ABM) - m( CBN) = 60^, so the triangle BMN is equilateral.