ABC is a triangle. The tangent to the circumcircle at A meets the line BC at D. The perpendicular to BC at B meets the perpendicular bisector of AB at E, and the perpendicular to BC at C meets the perpendicular bisector of AC at F. Show that D, E, F are collinear.
Solution
Let M be the midpoint of AB. Then ∠ BEM = ∠B, so BE = c/(2 sin B). Similarly, CF = b/(2 sin C). Hence BE/CF = (c sin C)/(b sin B).
∠DAB = ∠C, so ∠BDA = ∠B - ∠C. Hence BD = (c sin C)/sin(B-C). Similarly, ∠DAC = 180o - ∠B, so CD = (b sin B)/sin(B-C). Hence BD/CD = (c sin C)/(b sin B) = BE/CF. Hence triangles BED, CFD are similar, so ∠BDE = ∠BDF and so D, E, F are collinear.
Thanks to Alexandre Thiéry for this.
© John Scholes
jscholes@kalva.demon.co.uk
24 Nov 2003
Last corrected 24 Nov 03