Question

2. In ABC, the medians AD, BE and CF are concurrent and their point of concurrence
is G. Prove that ar (GAB) = ar (GBC) = ar (GCA) = 1/3 ar (ABC).
.
​

Answer 1

Given: A ∆ABC its medians AD, BE and CF intersect at G.

To prove : ar(∆AGB) = ar(∆AGC) = ar(∆BGC) = 1/3 ar(∆ABC).

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

In ∆ABC, AD is the median.

∴ ar(∆ABD) = ar(∆ACD) ...(i)

In ∆GBC, GD is the median.

∴ ar(∆GBD) = ar(∆GCD) ...(ii)

From (i) and (ii), we get

ar(∆ABD) - ar(∆GBD) = ar(∆ACD) -ar(∆GCD)

∴ a(∆AGB) = ar(∆AGC).

Similarly,

ar(∆AGB) =ar(∆AGC) =ar(∆BGC) ....(iii)

But, ar(ABC) = ar(∆AGB) + ar(∆AGC) + ar(∆BGC)

= 3 ar(∆AGB) [Using (iii)]

∴ ar(∆AGB) = 1/3 ar(∆ABC).

Hence, ar(∆AGB) = ar(∆AGC) = ar∆(BGC) = 1/3 ar(∆ABC).

I HOPE IT HELPS ☺️ FOLLOW ME DUDE (◕ᴗ◕✿)