ABC is a triangle. O is the circumcenter, D is the midpoint of AB, and E is the centroid of ACD. Prove that OE is perpendicular to CD iff AB = AC.
Solution
Use vectors. Take O as the origin. Let OA = A, OB = B, OC = C. Then OD = A/2 + B/2 and the midpoint of AD is 3/4 A + 1/4 B. Hence OE = 2/3 (3/4 A + 1/4 B) + 1/3 C = A/2 + B/2 + C/3, whilst CD = OD - OC = A/2 + B/2 - C. So OE is perpendicular to CD iff (A + B - 2C).(3A + B + 2C) = 0. Expanding and using A2 = B2 = C2, this becomes A.(B - C) = 0, which holds iff OA is perpendicular to BC. Hence result.
Many thanks to Fatih Olmez for the following geometrical solution:
Let K be the midpoint of CD, L the midpoint of AC and M the midpoint of BC. Let AM meet CD at G. G is the centroid, so GD = CD/3. K is the midpoint of CD, so KG = CD/6 and KG/DG = 1/2. E is the centroid of ACD, so KE/AE = 1/2. Hence GE is parallel to AD. But OD is perpendicular to AD, so it must be perpendicular to GE. In other words, G lies on the altitude from E in the triangle ODE.
If AB = AC, then O must lie on AM. So GO is the same as AO and is perpendicular to BC and hence to DL. In other words G lies on the altidue from O in the triangle ODE. Thus G is the orthocenter of ODE, so it must lie on the third altitude and hence OE is perpendicular to CD.
Conversely, if OE is perpendicular to CD (or equivalently to DG), then G lies on the altitude from D in the triangle ODE. Hence again it is the orthocenter and lies on the altitude from O. In other words, OG is perpendicular to DE (or DL) and hence to BC. OM is also perpendicular to BC (since O is the circumcenter), so G lies on OM. In other words, O, G, M are collinear. But A, G, M are collinear, so A lies on OM. In other words, A lies on the perpendicular bisector of BC, so AB = AC.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002