Question

If G is the centroid of a triangle ABC, then, prove analytically that
ABCG = ACAG =AABG.​

Answer 1

Step-by-step explanation:

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)

Answer 2

Answer:

Answer

If G is the centroid of the triangle △ABC

Then Area△AGB=Area△AGC=Area△CGB

Now Ar△AGB+Ar△AGC+Ar△CGB=Ar△ABC

∴3Ar△AGB=Ar△ABC

Hence, Ar△AGB=

3

1

Ar△ABC

Therefore, Ar△AGB=Ar△AGC=Ar△CGB=

3

1

ar△ABC