IMO 1990 SL 11

Problem

(IND 3′)IMO1 Given a circle with two chords AB, CD that meet at E, let M be a point of chord AB other than E. Draw the circle through D, E, and M. The tangent line to the circle DEM at E meets the lines BC, AC at F, G, respectively. Given AM AB = \lambda, find GE EF .

Solution

Assume B(A, E, M, B). Since A, B, C, D lie on a circle, we have \angleGCE = \angleMBD and \angleMAD = \angleFCE. Since FD is tangent to the circle around \triangleEMD at E, we have \angleMDE = \angleFEB = \angleAEG. Consequently, \angleCEF = 180◦−\angleCEA −\angleFEB = 180◦−\angleMED −\angleMDE = \angleEMD and \angleCEG = 180◦−\angleCEF = 180◦−\angleEMD = \angleDMB. It follows that \triangleCEF ∼\triangleAMD and \triangleCEG ∼ \triangleBMD. From the first similarity we obtain CE \cdot MD = AM \cdot EF, and from the second we obtain CE \cdot MD = BM \cdot EG. Hence AM \cdot EF = BM \cdot EG =⇒ GE EF = AM BM = \lambda 1 −\lambda . If B(A, M, E, B), interchanging the A B C D E F G M roles of A and B we similarly obtain GE EF = \lambda 1−\lambda.