Question
In a triangle $ABC,$ the medians $BE$ and $CF$ intersect at $G.$ Prove that $\text{ar}(\triangle\text{BCG})=\text{ar}(\text{AFGE}.)$
✓
Answer
Construction: Join $EF$
Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side, $FE || BC.$
Clearly, $\triangle\text{BEF}$ and $\triangle\text{CEF}$ are on the same base $EF$ and between the same parallel lines.
$\therefore\ \text{ar}(\triangle\text{BEF})=\text{ar}(\triangle\text{CEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BEF})-\text{ar}(\triangle\text{GEF})=\text{ar}(\triangle\text{CEF})-\text{ar}(\triangle\text{GEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BFG})=\text{ar}(\triangle\text{CEG})\dots(\text{i})$
We know that a median of a triangle divides it into two triangles of equal area.
$\therefore\ \text{ar}(\triangle\text{BEC})=\text{ar}(\triangle\text{ABE})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{CEG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{BFG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG}) [$Using $(i)]$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})=\text{ar(quadrilateral AFGE)}$
Since the line segment joining the mid-points of two sides of a triangle is parallel to the third side, $FE || BC.$
Clearly, $\triangle\text{BEF}$ and $\triangle\text{CEF}$ are on the same base $EF$ and between the same parallel lines.
$\therefore\ \text{ar}(\triangle\text{BEF})=\text{ar}(\triangle\text{CEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BEF})-\text{ar}(\triangle\text{GEF})=\text{ar}(\triangle\text{CEF})-\text{ar}(\triangle\text{GEF})$
$\Rightarrow\ \text{ar}(\triangle\text{BFG})=\text{ar}(\triangle\text{CEG})\dots(\text{i})$
We know that a median of a triangle divides it into two triangles of equal area.
$\therefore\ \text{ar}(\triangle\text{BEC})=\text{ar}(\triangle\text{ABE})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{CEG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG})$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})+\text{ar}(\triangle\text{BFG})$
$=\text{ar}(\text{quadrilateral AFGE})+\text{ar}(\triangle\text{BFG}) [$Using $(i)]$
$\Rightarrow\ \text{ar}(\triangle\text{BGC})=\text{ar(quadrilateral AFGE)}$
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