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Grade: 12th pass

                        

to prove ∆BCG=∆CAG=∆ABG,where G is the centroid of ∆ABC.

4 years ago

Answers : (2)

Saurabh Koranglekar
askIITians Faculty
8393 Points
							576-1246_1.PNG
						
one year ago
Tony
108 Points
							
We know that, the median of a triangle divide it into two triangles of equal area.
 
In ΔABC, AD is the median
 
∴ ar (ΔABD) = ar (ΔACD) ...(1)
 
In ΔGBC, GD is the median.
 
∴ ar(ΔGBD) = ar(ΔGCD) ...(2)
 
Subtracting (2) from (1), we get
 
ar(ΔABD) – ar(ΔGBD) = ar(ΔACD) – ar(ΔGCD)
 
∴ ar(ΔAGB) = ar(ΔAGC) ...(3)
 
Similarly, ar(ΔAGB) = ar(ΔBGC) ...(4)
 
From (3) and (4), we get
 
ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) ...(5)
 
Now, ar(ΔAGB) + ar(ΔAGC) + ar(ΔBGC) = ar(ΔABC)
 
 ar(ΔAGB) + ar(ΔAGB) + ar(ΔAGB) = ar(ΔABC) (Using (5))
 
 3ar(ΔAGB) = ar(ΔABC)
 
⇒ ar(ΔAGB) ... (6)
 
From (5) and (6), we get
 
ar(ΔAGB) = ar (ΔAGC) = ar(ΔBGC).
Hence it is proved.
Plzz approve if you like my proof! 
one year ago
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