Question

If the median of a ΔABC intersect at G, show that,
ar(ΔAGB) ar(ΔBGC)=ar(ΔAGC)=1/3(ΔABC)

Answer 1

Hey mate hope it helps you.....

Answer 2

Given: A ΔABC with medians AD, BE and CF intersecting at G.

To prove: ar(ΔAGB) = ar(Δ BGC) = ar(ΔAGC) =1/3 ar(ΔABC)

Proof: A median divides a triangle into two triangles of equal area.

In ΔABC, AD is the median.

⇒ ar(ΔABD) =ar(ΔADC)...(i)

In ΔGBC, Gd is the median.

⇒ ar(ΔGBD)= ar(ΔGDC)....(ii)

From the equations (i) and (ii), we get

ar(ΔABC) - ar(ΔGBD) = ar(ΔADC) - ar(ΔGDC)

⇒ ar(ΔAGB) = ar(ΔAGC)

Similarly, it can be also proved that,

ar(ΔAGB) = ar(ΔBGC)

⇒ ar(ΔAGB) = ar(ΔAGC) = ar(ΔBGC)

Now,

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

                = 3 ar(ΔAGB)

∴ ar(ΔAGB) = 1/3 ar(ΔABC)

⇒ ar(ΔAGB) ar(ΔBGC)=ar(ΔAGC)=1/3 ar(ΔABC) (is proved)