Two circles centers O and O' meet at A and B, so that OA is perpendicular to O'A. OO' meets the circles at C, E, D, F, so that the points C, O, E, D, O', F lie on the line in that order. BE meets the circle again at K and meets CA at M. BD meets the circle again at L and AF at N. Show that (KE/KM) (LN/LD) = (O'E/OD).
Solution
Solution by Milivoje Lukic
We have angle AOO' = 2 angle ACF, angle AO'O = 2 angle AFC. But angle AOO' + angle AO'O = 180o - angle OAO' = 90o, so angle ACF + angle AFC = 45o.
Angle CAL = angle CAO + angle OAO' + angle O'AL = angle ACF (OAC isosceles) + 90o + (90o - 1/2 angle AO'L) = angle ACF + 180o - angle ABL = angle ACF + 180o - angle ABD (same angle) = 180o (ACBD cyclic). So CAL is a straight line. Similarly, KAF is a straight line.
Applying Menelaus to the triangle CME and line KAF, we get (KE/KM) (AM/AC) (FC/FE) = 1. Applying Menelaus to the triangle FND and line LAC, we get (LD/LN) (AN/AF) (CF/CD) = 1. Hence (KE/KM) (LN/LD) = (AC/AM) (FE/FC) (AN/AF) (CF/CD) = (AC/AM) (AN/AF) (FE/CD) = (AC/AM) (AN/AF) (O'E/OD). So it suffices to show that AC/AM = AF/AN, or equivalently, that MN is parallel to CF.
Angle MAN = angle CAF = 180o - (angle ACF + angle AFC) = 135o. Angle MBN = angle EBA + angle ABD = angle EFA (AFBE cyclic) + angle ACD (ACBD cyclic) = 45o. So AMBN is cyclic, so angle AMN = angle ABN = angle ABD (same angle) = angle ACD (ACBD cyclic) = angle ACF (same angle). Hence MN parallel to CF as required.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002