Question

prove that medians of a triangle are concurrent

Answer 1

well !! I am proving this by using co-ordinate geometry .

Let ABC is a traingle in such a way that A(x1 , y1) , B(x2, y2) and C(x3, y3) .also Q is centoid of triangle . and AX, BY, and CZ are the midian of ∆ .

we know,
AX is median when , X is midpoint of BC
BY is median when, Y is midpoint of CA
CZ is median when, Z is midpoint of AB

use, section formula for finding points X, Y, and Z

point X = { (x2 + x3)/2, (y2+ y3)/2 }
Point Y = {(x1 + x3)/2 , (y1 + y3)/2}
Point Z = {(x1 + x2)/2 , (y1 + y2)/2 }

we also know, that Q( centroid ) divides median 2 : 1 ratio .

so,
AQ : QX = 2 :1
so,
point Q =[ { 2.(x2 + x3)/2 + 1.x1}/3 , {2.(y2 + y3)/2 + 1.y1 }/3 ]
= { (x1 + x2 + x3)/3 , (y1 +y2 + y3)/3 }

BQ : QY =2 :1
so,
point Q = [{2.(x1 + x3)/2 + 1.x2}/3 , {2.(y1 + y3)/2 + 1.y2 }/3 ]
= {(x 1 + x2 + x3)/3 , (y1 + y2 + y3)/3 }

CQ : QZ = 2 :1
so,
point Q =[ {2.(x1 + x2)/2 +1.x3}/3, {2.(y1 +y2)/2 + 1.y3 }/3 ]
= {(x1 + x2 + x3)/3 , (y1 + y2 +y3)/3 }

Here, we see in all cases point Q always same .e.g Q = { (x1 + x2 + x3)/3 , (y1 + y2 + y3)/3} .
hence, Q is one and only point from where all medians are passing .

hence, medians of ∆ are concurrent

Answer 2

hope this will help you