Concurrent Lines


   
 
Theorem 4
Statement
 
The medians of a triangle pass through the same point which divides each of the medians in the ratio 2:1.
 
 
Given:
 
In DABC, medians BE and CF intersects at G. AG is joined and produced to meet BC in D.
 
To prove:
 
i) AD is the median.
 
i.e., BD = DC
 
ii) AG : GD = BG : GE = CG : GF = 2 : 1
 
Construction:
 
 
Produce AD to K such that AG = GK. Join KB and KC.
 
Proof:
 
i) AD is the median
 
i.e., BD = DC
 
In DABK,
 
F is the mid-point of AB . (given)
 
G is the mid-point of AK. (by construction)
 
FG || BK
 
or GC || BK ...(1)
 
By theorem on line joining the mid-points of any two
 
sides of a triangle.
 
Similarly in DACK,
 
GE || CK or GB || CK
 
 
From (1) and (2), BKCG is a parallelogram.
 
Now in BKCG, BC and GK are diagonals.
 
The diagonals of a parallelogram bisect each other.
 
BD = DC or AD is the median from A.
 
(ii) GK and BC are diagonals of GCKB.
 
GD = DK
 
 
 
 
 
i.e., AG : GD = 2 : 1
 
Similarly, BG : GE = 2 : 1 and
 
CG : GF = 2 : 1
 
Hence the theorem is proved.
 
 
     
   
Get FREE Live Tutoring
Get FREE Live Tutoring
(No credit card required)

Customer Care

Click to get customer service, technical support and subscription help.

Customer Care Chat


Refer-A-Friend

Get One Month Free!
When you refer a friend