In a triangle ABC, position vectors of A,B,C are a,b,c respectively. D is the midpoint of BC, and G is the centroid. (a) Find AD in terms of a,b,c. (b) Find AG. (c) Show that G divides AD in the ratio 2:1.
Solution — Step by Step
D is the midpoint of BC:
d=2b+cAD=d−a=2b+c−a=2b+c−2a
The centroid is the average of the three vertices:
g=3a+b+cAG=g−a=3a+b+c−a=3b+c−2a
Compare:
ADAG=(b+c−2a)/2(b+c−2a)/3=32
So G lies on AD at 32 of the way from A. That means AG=32AD and GD=31AD, giving AG:GD=2:1.
AD=(b+c−2a)/2, AG=(b+c−2a)/3, ratio AG:GD=2:1.
Why This Works
Position vectors turn geometry into algebra. Midpoints become averages of two vectors; centroids become averages of three. Ratios and concurrencies follow from simple vector arithmetic — no construction needed.
The fact that medians are concurrent at G, and divide it in 2:1, is a direct consequence of g=(a+b+c)/3.
Alternative Method
Section formula approach: since G divides AD in some ratio k:1, write g=(kd+a)/(k+1) and equate to (a+b+c)/3. Solving gives k=2. Same answer, more algebra.
Common Mistake
Writing the centroid as (a+b+c)/2 instead of dividing by 3. The factor 3 reflects three vertices. A common slip when students confuse it with the midpoint formula.
Want to master this topic?
Read the complete guide with more examples and exam tips.