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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