If the medians of a ΔABC intersect at G. Prove that ar(ΔBGC) is equal to 13ar(ΔABC).
Answer:
- We are given that AD, BE and CF are the medians of ΔABC intersecting at G.
Now, we have to find the area of ΔBGC. - We know that a median of a triangle divides it into two triangles of equal area.
Now, in ΔABC, AD is the median. ∴ Similarly, in \Delta GBC, GD is the median. \begin{aligned} \therefore ar(\Delta GBD)= ar(\Delta GCD) &&\ldots\text{(ii)} \end{aligned} - From \text{(i)} and \text{(ii)}, we get: \begin{aligned} &ar(\Delta ABD) - ar(\Delta GBD)= ar(\Delta ACD) - ar(\Delta GCD) \\ \implies& ar(\Delta AGB) = ar(\Delta AGC) \end{aligned} Similarly, \begin{aligned} & ar(\Delta AGB) = ar(\Delta BGC) \\ \therefore \space & ar(\Delta AGB) = ar(\Delta AGC) = ar(\Delta BGC) &&\ldots\text{(iii)} \end{aligned}
- \begin{aligned} & \text{But,}\\ & ar(\Delta ABC) = ar(\Delta AGB) + ar(\Delta AGC) + ar(\Delta BGC) = 3 \space ar(\Delta BGC) &&[ \text{Using eq (iii)} ] \\ &\therefore ar(\Delta BGC) = \dfrac { 1 } { 3 } ar(\Delta ABC) \end{aligned}