Let us consider a segment #PQ# (shown below in Fig.1), which is divided by a point #R# in the ratio of #l:m#. Then vector representing #R# is given by #(mvecp+lvecq)/(l+m)#

It is apparent that mid point is represented by #(vecp+vecq)/2#.

Now let us consider the #DeltaABC#, where #A,B# and #C# are reprsented by #vecA,vecB# and #vecC# respectively. #D,E# and #F# are the midpoints of #BC,AC# and #AB# respectively. Let the vectors representing #D,E# and #F# be #vecd.vece# and #vecf#. #AD,BE# and #CF# are joined to intersect at #G# represented by #vecg#.

It is aparent that #vecd=(vecb+vecc)/2#, #vece=(veca+vecc)/2# and #vecf=(veca+vecb)/2#.

Let us also find the #vecg#. As #G# divides #AD# in the ratio of #2:1#, we have

#vecg=(2vecd+veca)/(2+1)=(2vecd+veca)/(2+1)=(2((vecb+vecc)/2)+veca)/3=(veca+vecb+vecc)/3#

Observe that it is symmetric w.r.t. the median chosen and had we divided #BE# or #CF# in theratio of #2:1#, the result would have been same. And #G# is the centroid of #DeltaABC#.

Hence, medians of a triangle are concurrent.

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