ABC is an acute-angled triangle which is not isosceles. M is the midpoint of BC. X is any point on the segment AM. Y is the foot of the perpendicular from X to BC. Z is any point on the segment XY. U and V are the feet of the perpendiculars from Z to AB and AC. Show that the bisectors of angles UZV and UXV are parallel.
Solution
Let the bisector of UZV meet the line AC at K. In the quadrilateral AUZV, angles AUZ and AVZ are 90o, so angle UZV is 180o - A. Hence ∠KZV = 90o - A/2. Now consider triangle KVZ. ∠KVZ = 90o, so ∠ZKV = A/2, So ZK is parallel to the bisector of angle A.
Let the line through X perpendicular to XY meet AB at P and AC at Q. Then ∠PXZ = ∠PUZ = 90o, so PUXZ is cyclic. Hence ∠XUZ = ∠XPZ. Similarly ∠XVZ = ∠XQZ. But XPZ and XQZ are congruent triangles, so ∠XUZ = ∠XVZ. Let XP' be the perpendicular from X to AB and XQ' the perpendicular from X to AC. Then angle P'XU = angle XUZ and ∠Q'ZV = ∠XVZ, so ∠P'XU = ∠Q'ZV. Hence the bisector of UXV also bisects P'XQ'. But the same argument as above shows that the bisector of P'XQ' is parallel to the bisector of angle A.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002