IMO 2000 SL G6

Origin: ARG | Category: Geometry

Problem

Let ABCD be a convex quadrilateral with AB not parallel to CD, let X be a point inside ABCD such that ∡ADX = ∡BCX < 90◦ and ∡DAX = ∡CBX < 90◦. If Y is the point of intersection of the perpendicular bisectors of AB and CD, prove that ∡AY B = 2∡ADX.

Solution

By the condition of the problem, \triangleADX and \triangleBCX are similar. Then there exist points Y ′ and Z′ on the perpendicular bisector of AB such that \triangleAY ′Z′ is similar and oriented the same as \triangleADX, and \triangleBY ′Z′ is (being congruent to \triangleAY ′Z′) similar and oriented the same as \triangleBCX. Since then AD/AY ′ = AX/AZ′ and \angleDAY ′ = \angleXAZ′, \triangleADY ′ and \triangleAXZ′ are also similar, implying AD AX = DY ′ XZ′ . Analogously, BC BX = CY ′ XZ′ . It follows from AD AX = BC BX that CY ′ = DY ′, which means that Y ′ lies on the perpendicular bisector of CD. Hence Y ′ \equivY .

Now \angleAY B = 2\angleAY Z′ = 2\angleADX, as desired.