Prove that medians of a triangle are concurrent

#### Solution

Consider ∆ABC.

Let P, Q, R be the midpoints of the sides BC, CA, AB respectively.

Let the medians BQ and CR intersect at G.

To prove that the third median AP also passes through G.

Let `bar"a", bar"b", bar"c", bar"p", bar"q", bar"r", bar"g"` be the position vectors of the points A, B, C, P, Q, R, G respectively.

Since P, Q, R are the mid-points of the sides BC, CA, AB respectively

∴ By midpoint formula, we get

`bar"p" = (bar"b" + bar"c")/2` .......(i)

`bar"q" = (bar"c" + bar"a")/2` .......(ii)

`bar"r" = (bar"a" + bar"b")/2` .......(iii)

From (i), (ii) and (iii), we get

`2bar"p" =bar"b" + bar"c" ⇒ 2bar"p" + bar"a" = bar"a" + bar"b" + bar"c"`

`2bar"q" = bar"c" + bar"a" ⇒ 2bar"q" + bar"b" = bar"a" + bar"b" + bar"c"`

`2bar"r" = bar"a" + bar"b" ⇒ 2bar"r" + bar"c" = bar"a" + bar"b" + bar"c"`

∴ `(2"p" + bar"a")/3 = (2bar"q" + bar"b")/3 = (2bar"r" + bar"c")/3 = (bar"a" + bar"b" + bar"c")/3`

∴ `(2"p" + bar"a")/(2 + 1) = (2bar"q" + bar"b")/(2 + 1) = (2bar"r" + bar"c")/(2 +1) = (bar"a" + bar"b" + bar"c")/3`

= `bar"g"` .......(say)

This shows that the point G whose position vector is `bar"g"` lies on the three medians AP, BQ, CR dividing them internally in the ratio 2:1.

Hence, the three medians are concurrent.