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.
 
  
     
   
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