prove by vector method that the medians of a triangle are concurrent.
Answers
Step-by-step explanation:
ANSWER
Let A,B and C be vertices of a triangle.
Let D,E and F be the midpoints of the sides BC,AC and AB respectively. Let
OA
=
a
,
OB
=
b
,
OC
=
c
,
OD
=
d
=
OE
=
e
and
OF
=
f
be position vectors of points A,B,C,D,E and F respectively.
Therefore, by Midpoint formula,
d
=
2
b
+
c
,
e
=
2
a
+
c
and
f
=
2
a
+
b
∴2
d
=
b
+
c
,2
e
=
a
+
c
and 2
f
=
a
+
b
∴2
d
+
a
=
a
+
b
+
c
,
2
e
+
b
=
a
+
b
+
c
,
2
e
+
b
=
a
+
b
+
c
2
f
+
c
=
a
+
b
+
c
Now,
3
2
d
+
a
=
3
2
e
+
b
=
3
2
f
+
c
=
3
a
+
b
+
c
Let
g
=
3
a
+
b
+
c
. Then, we have
g
=
3
a
+
b
+
c
=
2+1
(2)
d
+(1)
a
=
2+1
(2)
e
+(1)
b
=
2+1
(2)
f
+(1)
c
If G is the point whose position vector is
g
, then from the above equation it is clear that the point G lies on the medians AB,BE,CF and it divides them internally in the ratio 2:1.
Hence, the medians of a triangle are concurrent.