Question

Question 5 (v)
In the following figure, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:

(v) ar(BFE) = 2ar(FED)

Answer 1

Let ‘h’ be the height of vertex E, corresponding to the side BD in $\Delta BDE.$
Let ‘H’ be the height of vertex A, corresponding to the side BC in $\Delta ABC$.
In (i) , It was shown that
i.e., $Ar\left(BDE\right)=\frac{1}{4}ar\left(ABC\right)$
$\therefore \frac{1}{2}\times BD\times h=\frac{1}{4}(\frac{1}{2}\times BC\times H)$
$\Rightarrow BD\times h=\frac{1}{4}(2BD\times H)$
$\Rightarrow BD\times h=\frac{1}{4}(2BD\times H)$
$\Rightarrow h=\frac{1}{2}H$
In (iv), it was shown that $ar\left(\Delta BFE\right)=ar\left(\Delta AFD\right)$.
i.e., $ar\left(\Delta BFE\right)=ar\left(\Delta AFD\right)$
$=\frac{1}{2}\times FD\times H=\frac{1}{2}\times FD\times 2h=2(\frac{1}{2}\times FD\times h)$
$=2ar\left(\Delta FED\right)$
Hence, ar (BFE) = 2ar (FED)