Question

The vertex A of ∆ABC is joined to a point D on BC. If E is the midpoint of AD, then ar(∆BEC) = ?
(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$
(b) $\frac{1}{3}\mathrm{ar}\left(∆ABC\right)$
(c) $\frac{1}{4}\mathrm{ar}\left(∆ABC\right)$
(d) $\frac{1}{6}\mathrm{ar}\left(∆ABC\right)$

Answer 1

(a) $\frac{1}{2}\mathrm{ar}\left(∆ABC\right)$

Since E is the midpoint of AD, BE is a 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)
Since E is the midpoint of AD, CE is a median of ∆ADC.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆CED ) = $\frac{1}{2}$⨯ ar(∆ADC) ...(ii)

Adding (i) and (ii), we have:
ar(∆BED ) + ar(∆CED ) = $\frac{1}{2}$ ⨯ ar(∆ABD) + $\frac{1}{2}$⨯ ar(∆ADC)
⇒ ar (∆ BEC ) = $\frac{1}{2}\left(∆ABD+ADC\right)=\frac{1}{2}∆ABC$