Class 9MathematicsAreas of Parallelograms and TrianglesMix Examples - Areas of Parallelograms and Triangles
MediumIn $\Delta ABC$,medians $AD$,$BE$,and $CF$ intersect at point $G$. Prove that,$ar(GAB) = ar(GBC) = ar(GCA) = \frac{1}{3} ar(ABC)$.
(N/A) $1$. In $\Delta ABC$,$AD$ is the median. Since a median divides a triangle into two triangles of equal area,$ar(ABD) = ar(ACD) = \frac{1}{2} ar(ABC)$.
$2$. Similarly,in $\Delta GBC$,$GD$ is the median,so $ar(GBD) = ar(GCD)$.
$3$. Subtracting these areas from the larger triangles: $ar(GAB) = ar(ABD) - ar(GBD)$ and $ar(GAC) = ar(ACD) - ar(GCD)$. Since $ar(ABD) = ar(ACD)$ and $ar(GBD) = ar(GCD)$,it follows that $ar(GAB) = ar(GAC)$.
$4$. By symmetry,using medians $BE$ or $CF$,we can show $ar(GAB) = ar(GBC) = ar(GCA)$.
$5$. Since the sum of these three areas is $ar(ABC)$,each must be equal to $\frac{1}{3} ar(ABC)$.
$2$. Similarly,in $\Delta GBC$,$GD$ is the median,so $ar(GBD) = ar(GCD)$.
$3$. Subtracting these areas from the larger triangles: $ar(GAB) = ar(ABD) - ar(GBD)$ and $ar(GAC) = ar(ACD) - ar(GCD)$. Since $ar(ABD) = ar(ACD)$ and $ar(GBD) = ar(GCD)$,it follows that $ar(GAB) = ar(GAC)$.
$4$. By symmetry,using medians $BE$ or $CF$,we can show $ar(GAB) = ar(GBC) = ar(GCA)$.
$5$. Since the sum of these three areas is $ar(ABC)$,each must be equal to $\frac{1}{3} ar(ABC)$.
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