If the medians of a ^@\Delta ABC^@ intersect at ^@G^@. Prove that ^@ar(\Delta BGC) ^@ is equal to ^@ \dfrac { 1 } { 3 } ar(\Delta ABC) ^@.
Answer:
- 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^@. - 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} @^ - 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} @^
