Question

Question 5 (vi)
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:
$ar\left(FED\right)=\frac{1}{8}ar\left(AFC\right)$

Answer 1

Area (AFC) = area (AFD) + area (ADC)
$=ar\left(BEF\right)+\frac{1}{2}ar\left(ABC\right)\left[In\right(iv),ar(BFE)=ar(AFD);ADismedianof\Delta ABC]$
$=ar\left(BFE\right)+\frac{1}{2}\times 4ar\left(BDE\right)\left[In\right(i),ar(BDE)=\frac{1}{4}ar(ABC\left)\right]$
= ar(BEF) + 2ar(BDE)....(5)
Now, by (v),
ar(BFE) = 2ar(FED) …….(6)
ar(BDE) = ar(BFE) + ar(FED)
= 2ar(FED) + ar(FED)
= 3ar(FED) ……(7)
Therefore, from equations (5), (6), and (7), we get:
$ar\left(AFC\right)=2ar\left(FED\right)+2\times 3ar\left(FED\right)=8ar\left(FED\right)$
$\therefore ar\left(AFC\right)=8ar\left(FED\right)$
Hence, $ar\left(FED\right)=\frac{1}{8}ar\left(AFC\right)$