74th Romanian Mathematical Olympiad

Question

Let ABC be a triangle with BAC = 30^ and ABC = 100^. Let m be the perpendicular bisector of AC, E be the intersection of m and AB, and D be the point on m, inside the triangle ABC, such that CAD = 10^. Let M be the intersection of the lines AD and CE. a) Prove that CE is the bisector of the angle BCD. b) Prove that AM = AB. FIGURE_0

Accepted Answer

a) As m is the perpendicular bisector, we have DA = DC and EA = EC. Thus DEA DEC. FIGURE_0 From DA = DC, we get DCA = DAC = 10^. We have DCE = DAE = 20^, so BCE = 20^ and CE is the bisector of the angle BCD. b) As CDA = 160^, it follows ADE = CDE = 100^ = CBE. We obtain CEB = CED = 60^, so CEB CED (A.S.A.) It follows that BE = BD. But MEB = MED and ME = ME, so MEB MED (S.A.S.). We infer that EBM = EDM = 180^ - CED - AME = 180^ - 60^ - 40^ = 80^. But BAM = 20^, so AB = AM.