Question

The vertex A of $△ABC$ is joined to a 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

Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: ar(ΔBEC)=ar(ΔABC)
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 ) = ar(∆ABD) ...(i)
Also, ar(∆CDE ) = ar(∆ADC) ...(ii)

From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE)​ = ⨯​ ar(∆ABD)​ + ⨯​ ar(∆ADC)
⇒ ar(∆BEC )​ = ⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ​ar(∆BEC )​ =​ ⨯​ ar(∆ABC)