Question

The vertex $A$ of $△ABC$ is joined to point $D$ on the side $BC$. The midpoint of $AD$ is $E$. Prove that $ar\left(△BEC\right)=\frac{1}{2}ar\left(△ABC\right)$.

Answer 1

From the given figure we know that $BE$ is the median of $△ABD$

So we get

Area of $△BDE$= Area of $△ABE$

It can be written as

Area of $△BDE=\frac{1}{2}$ ( Area of $△ABD$).....(1)

From the figure we know that $CE$ is the median of $△ADC$

So we get

Area of $△CDE$ = Area of $△ACE$

It can be written as

Area of $△CDE=\frac{1}{2}$ ( Area of $△ACD$)......(2)

By adding both the equations

Area of $△BDE$+ Area of $△CDE=\frac{1}{2}$ (Area of $△ABD)+\frac{1}{2}$ ( Area of $△ACD$)

By taking $\frac{1}{2}$ as common

Area of $△BEC=\frac{1}{2}$ ( Area of $△ABD$+ Area of $△ACD)$

So we get

Area of $△BEC=\frac{1}{2}$ ( Area of $△ABC$)

Therefore, it is proved that $ar\left(△BEC\right)=\frac{1}{2}ar\left(△ABC\right)$.