ABC is a triangle with AB and AC unequal. AM, AX, AD are the median, angle bisector and altitude. Show that X always lies between D and M, and that if the triangle is acute-angled, then angle MAX < angle DAX.
Solution
wlog ∠B > ∠C. So ∠B + ∠B + ∠A > ∠C + ∠B + ∠A = 180o. Hence ∠B + ∠A/2 > 90o. Hence ∠AXB = 180o - ∠B - ∠A/2 < 90o. So D must lie inside the segment BX. Since ∠B > ∠C, we have AC > AB. But CX/BX = AC/AB, so CX > BX. Hence M lies inside the segment CX. So X lies between D and M.
Let O be the circumcenter and N the midpoint of the arc BC. Then A, X, N are collinear. Since the triangle is acute-angled, O must lie on the opposite side of BC to N, so ∠MAN < ∠OAN = ∠DAN.
© John Scholes
jscholes@kalva.demon.co.uk
1 Nov 2003
Last corrected/updated 1 Nov 03