Question

IF AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in area. If G is the midpoint of median AD, prove that ar $\left(△BGC\right)=2ar\left(△AGC\right).$

Answer 1

Given: In \triangle ABC, AD is its median. G is the midpoint of AD. BG and CF are joined.

To prove:
(i) ar $\left(△ADB\right)=ar\left(△ADC\right)$
(ii) ar $\left(△BGC\right)=2ar\left(△AGC\right)$
Construction : $DrawAL\perp BC$

Proof : (i) $\because$ AD is the median of $△ABC$
$\therefore BD=DC$
Now $ar\left(△ABD\right)=\frac{1}{2}base\times altitude$
$=\frac{1}{2}BD\times AL$
and ar $\left(△ACD\right)=\frac{1}{2}\times CD\times AL=\frac{1}{2}\times BD\times AL$
From (i) and (ii),
ar $\left(△ABD\right)=ar\left(△ACD\right)$
In $△BGC,GD$ is the median
$\therefore ar\left(△BGD\right)=ar\left(△CGD\right)$
Similarly in $△ACD,$ G is the midpoint of AD
$\therefore CG$ is the median
$\therefore ar\left(△AGC\right)$ = $ar\left(△CGD\right)$
From (i) and (ii) ,
ar $\left(△BGD\right)$ = ar $\left(△AGC\right)$
But ar $\left(△BGC\right)$ = $2ar\left(△BGD\right)$
$\therefore ar\left(△BGC\right)=2ar\left(△AGC\right)$