### If the medians of a $\Delta ABC$ intersect at $G$. Prove that $ar(\Delta BGC)$ is equal to $\dfrac { 1 } { 3 } ar(\Delta ABC)$. A C B D E F G

Step by Step Explanation:
1. We are given that $AD$, $BE$ and $CF$ are the medians of $\Delta ABC$ intersecting at $G$.

Now, we have to find the area of $\Delta BGC$.
2. We know that a median of a triangle divides it into two triangles of equal area.

Now, in $\Delta ABC$, $AD$ is the median. \begin{aligned} \therefore ar(\Delta ABD)= ar(\Delta ACD) &&\ldots\text{(i)} \end{aligned} Similarly, in $\Delta GBC$, $GD$ is the median. \begin{aligned} \therefore ar(\Delta GBD)= ar(\Delta GCD) &&\ldots\text{(ii)} \end{aligned}
3. 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}
4. \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}