Question

Prove that median of a triangle intersect in ratio 2:1.
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Answer 1

Here we assume a traingle ABC such that O is the centroid and E is the median of side BC , F of side AC , G of side AB. Vertice A is represented by vector a , B by vector B , C by vector c.

To prove : the median of a triangle intersect each other in the ratio 2:1

as we know by mis point theorm , the E would be represented by = (vector b + vector c) /2 {mid point of side ................(1) BC}

F would be represented by = (vector a + vector c) /2

{mid point of side BC}

G would be represented by = (vector a + vector b) /2

{mid point of side AB}

as we know the centroid is represented by the sum of vectors of vertices divided by no. of sides i.e 3

therefore O will be represented as

= (vector a + vector b + vector c ) /3.........(2)

now as we know by section formula , any side divided by the ration m : n will be represented by

=( m (vector b) + n (vector a) ) / (m +n)

applying the above on the side AE divided by the centroid O in ratio m :n

we get ,

=> (vector a + vector b + vector c) /3 = m ( vector b + vector c ) /2 ) + n( vector a)

{ using (1) and (2) }

=> (vector a + vector b + vector c) /3 = ( m (vector b ) + m ( vector c ) + 2 n ( vector a) ) / [2 (m+n) ]

after comparing like values we get n = 1/2 , m = 1

therefore the required ratio m: n = 1 / ( 1/2 ) = 2 :1

this implies that the median of a triangle intersect each other in the ratio 2:1 .

Hence Proved