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If $AD$ is a median of a triangle $ABC$, then prove that triangles $ADB$ and $ADC$ are equal in area. If $G$ is the mid point of median $AD$, prove that $ar(\triangle BGC) = 2ar(\triangle AGC)$.
Given:
$AD$ is a median of a triangle $ABC$.
$G$ is the mid point of median $AD$.
To do:
We have to prove that triangles $ADB$ and $ADC$ are equal in area and $ar(\triangle BGC) = 2ar(\triangle AGC)$.
Solution:
Join $BG$ and $CF$.
Draw $AL \perp BC$
$\mathrm{AD}$ is the median of $\triangle \mathrm{ABC}$
This implies,
$\mathrm{BD}=\mathrm{DC}$
$\operatorname{ar}(\triangle \mathrm{ABD})=\frac{1}{2}$ base $\times$ altitude
$=\frac{1}{2} \mathrm{BD} \times \mathrm{AL}$.........(i)
$\operatorname{ar}(\triangle \mathrm{ACD})=\frac{1}{2} \times \mathrm{CD} \times \mathrm{AL}$
$=\frac{1}{2} \times \mathrm{BD} \times \mathrm{AL}$.............(ii) (Since $\mathrm{BD}=\mathrm{DC}$)
From (i) and (ii),
$\operatorname{ar}(\triangle \mathrm{ABD})=\operatorname{ar}(\triangle \mathrm{ACD})$
In $\triangle \mathrm{BGC}, \mathrm{GD}$ is the median.
$\operatorname{ar}(\triangle \mathrm{BGD})=\operatorname{ar}(\triangle \mathrm{CGD})$
Similarly,
In $\triangle \mathrm{ACD}, \mathrm{G}$ is the mid point of $\mathrm{AD}$, $\mathrm{CG}$ is the median.
$\operatorname{ar}(Delta \mathrm{AGC})=\operatorname{ar}(\Delta \mathrm{CGD})$
From (i) and (ii),
$\operatorname{ar}(\triangle \mathrm{BGD})=\operatorname{ar}(\triangle \mathrm{AGC})$
$\operatorname{ar}(\triangle \mathrm{BGC})=2 \operatorname{ar}(\Delta \mathrm{BGD})$
This implies,
$\operatorname{ar}(\Delta \mathrm{BGC})=2 \operatorname{ar}(\triangle \mathrm{AGC})$
Hence proved.
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