mathlinE lesson  Copyright 2004 mathlinE
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Proofs of concurrency of (a) altitudes
(b) perpendicular bisectors (c) medians of any
triangle by vector methods. (Suitable for years 11, 12)
Preliminaries:
(1) Two non-zero vectors are perpendicular if their
dot product equals zero.
(2) If pa + qb = 0, and a and b are independent
vectors, then p = q = 0.
(3) If M is the mid-point of AB , then
1
OM  (a + b).
A
M
2
B
a
b
Concurrency of medians of any triangle
Proof:
B
O
Concurrency of altitudes of any triangle
Proof:
B
P b
O
a
Let OA  a, OB  b and OC  c.  AC  c – a,
BC  c – b and BA  a – b. Since M, N and P are
1
1
mid-points, OM  (c + b), ON  (c + a) and
2
2
1
OP  (a + b). Since OM and ON are
2
perpendicular to BC and AC respectively,
1
 OM  BC  (c + b).(c – b) = 0 and
2
1
ON  AC  (c + a).(c – a) = 0.
2
Hence |c|2 – |b|2 = 0 and |c|2 – |a|2 = 0,  |a| = |b| = |c|.
1
1
OP  BA  (a + b).(a – b) = [ |a|2 – |b|2 ] = 0,
2
2
OP is perpendicular to BA ,  it is a perpendicular
bisector of BA . Hence the three perpendicular
bisectors are concurrent.
P
M
b
O
a
A
c
A
N
C
AM and BN are altitudes and CP is a line segment
passing through the intersection O of AM and BN .
Let OA  a, OB  b and OC  c.  AC  c – a and
b.(c – a) = 0, BC  c – b and a. (c – b) = 0.
Hence b.c – b.a = 0 and a.c – a.b = 0,  b.c – a.c = 0,
or (b – a).c = 0, i.e. c is perpendicular to AB and
hence CP is perpendicular to AB , i.e. CP is an
altitude. Hence the three altitudes are concurrent. A
similar proof can be constructed for an obtuse
triangle. Try it as an exercise.
Concurrency of perpendicular bisectors of any 
Proof:
B
P
M
b
O
a
c
A
N
C
OM and ON are perpendicular bisectors of BC and
AC respectively, and OP bisects AB .
M
c
C
N
AM and BN are medians and CP is a line segment
passing through the intersection O of AM and BN .
1
Let OA  a, OB  b and OC  c. OM  (b + c)
2
1
and ON  (a + c). Let m, n and p be some positive
2
constants such that OM  m a, ON  n b and
1
1
OP   p c. m a  (b + c) and  n b  (a + c).
2
2
From the last two equations, –2ma – b = –2nb – a,
hence (1 – 2m)a – (1 – 2n)b = 0. Since a and b are not
parallel,  they are independent and hence 1 – 2m = 0
1
1
1
and –(1 – 2n) = 0, i.e. m = n = .   b  (a + c)
2
2
2
and b = –a – c. AP  AO  OP  –a – pc, and
AB  b – a = –2a – c. Let AP  k AB where k is a
positive constant.  –a – pc = k(–2a – c) or
(2k – 1)a + (k – p)c = 0. Since a and c are
1
independent, 2k  1  0 and k  p  0 , i.e. k 
2
1
and p  .  P is the mid-point of AB and CP is a
2
median. Hence the three medians are concurrent and
the intersection trisects each one.