Question

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

Answer 1

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: $\mathrm{ar}(∆BEC)=\frac{1}{2}\mathrm{ar}(∆ABC)$
Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = $\frac{1}{2}$ar(∆ABD) ...(i)
Also, ar(∆CDE ) =$\frac{1}{2}$ ar(∆ADC) ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ = $\frac{1}{2}$ ⨯​ ar(∆ABD)​ + $\frac{1}{2}$ ⨯​ ar(∆ADC)
⇒ ar(∆BEC )​ = $\frac{1}{2}$⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ​ar(∆BEC )​ =​ $\frac{1}{2}$ ⨯​ ar(∆ABC)