if the medians of a triangle ABC intersect at G show that ar ( AGB ) = ar(AGC)=ar(BGC)=1/3 ar(ABC).

Dear Student!

Here is the answer to your query.

 

Given: In ΔABC, the medians AD, BE and CF intersect in G.

To Prove: ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC) = ar(ΔABC)

Proof: 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) = ar(ΔABC)    ...(6)

From (5) and (6), we get

ar(ΔAGB) = ar (ΔAGC) = ar(ΔBGC) = ar(ΔABC)

Cheers!

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