In fig, $ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. If $AE$ intersects $BC$ at $F$, show that:
(i) $ar(BDE)=\frac{1}{4}ar(ABC)$

(ii) $ar(BDE)=\frac{1}{2}ar(BAE)$

(iii) $ar(ABC)=2ar(BEC)$

(iv) $ar(BFE)=ar(AFD)$

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

(vi) $ar(FED)=\frac{1}{8}ar(AFC)$

Given : $ABE$ and $BDE$ are two equilateral triangles.

Considering $a$ as the side of $△ABC.$

Then, ar$(ABC)=\frac{\sqrt{3}}{4}{a}^2$

(i)  Since $D$ is the midpoint of $BC.$  

So, $BD=\frac{a}{2}$

ar$(BDE)=\frac{\sqrt{3}}{4}{\left(\frac{a}{2}\right)}^2=\frac{\sqrt{3}}{16}{a}^2$ 

$\Rightarrow ar(BDE)=\frac{1}{4}ar(ABC)$

(ii) Since $DE$ is a median of $△BEC$ and each median divides a triangle in two other  $△$s of equal area, 

So, ar$(BDE)=\frac{1}{2}ar(BEC)$ ... (1) 

$\angle EBC=\angle BCA=60^o$ 

This suggests that they are alterante angles between the lines $BE$ and $AC$ with $BC$ as the transversal.

So, $BE\parallel AC.$ 

Since $△BEC$ and $BAE$ stand on the same base $BE$ and lie between the same parallel lines $BE$ and $AC.$ 

So, $ar(BEC)=ar(BAE)$

$ar(BDE)=\frac{1}{2}ar(BEC)$ (from (1))

ar $(\Delta BDE)=\frac{1}{2}ar(BAE).$ 

(iii) $ar(BDE)=\frac{1}{2}ar(BEC)$ 

($ED$ is a median of $△BEC$ and median divides the triangle in two other $△s$ of equal area. )

$ar(BDE)=\frac{1}{4}ar(ABC)$  (Using result from (i))

So, $\frac{1}{4}ar(ABC)=\frac{1}{2}ar(BEC)$

$\Rightarrow ar(ABC)=2ar(BEC)$

(iv) $\angle ABD=\angle BDE=60^{\circ }$ (Angles of equilateral triangle)

These are alternate angles between the lines $AB$ and $ED$ with $BD$ as transversal. 

So, $BA$ || $ED.$ 

So, $ar(BDE)=ar(AED)$ ( Δs  on the same base $ED$ and between the same parallel lines $AB$ and $DE$) 

Subtracting $ar(FED)$ from both sides,

$\Rightarrow ar(BDE)–ar(FED)=ar(AED)–ar(FED)$

$\Rightarrow ar(BEF)=ar(AFD)$

(v) In  Δ  $ABC,$

By Appolonius theorem,   

$A{B}^2+A{C}^2=2A{D}^2+2B{D}^2$

Since $AB=AC,$

$A{D}^2=A{B}^2-B{D}^2$

$A{D}^2=(a{)}^2-{\left(\frac{a}{2}\right)}^2$

$A{D}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4}$

$AD=\frac{\sqrt{3}}{2}a$

In  Δ $BED,$  

By using Appolonius theorem,

Drop perpendicular from $EP$ to $BD,$

So , $EP$ is the median $BED$

$E{P}^2=D{E}^2-D{P}^2$

$E{P}^2={\left(\frac{a}{2}\right)}^2-{\left(\frac{a}{4}\right)}^2$

$E{P}^2=\frac{a^2}{4}-\frac{a^2}{16}=\frac{3a^2}{16}$  

$\Rightarrow EP=\frac{\sqrt{3}a}{4}$

$ar(AFD)=\frac{1}{2}\times FD\times AD$

$ar(AFD)=\frac{1}{2}\times FD\times \frac{\sqrt{3}}{2}a$  

$ar(EFD)=\frac{1}{2}\times FD\times EP$

$ar(EFD)=\frac{\sqrt{3}}{4}a$

Now, 

$ar(AFD)=\frac{\sqrt{3}}{2}a$

$ar(EFD)=\frac{\sqrt{3}}{4}a$

$ar(AFD)=2ar(EFD)$ 

Also, $ar(BEF)=ar(AFD)$

So, $ar(BFE)=ar(AFD)=2ar(EFD)$

(vi) $ar(BDE)=\frac{1}{4}ar(ABC)$

So, $\Rightarrow ar(BEF)+ar(FED)=\frac{1}{4}\times 2ar(ADC)$

$2ar(FED)+ar(FED)=\frac{1}{2}(ar(AFC)–ar(AFD)$ [Using part (v)]

$3ar(FED)=\frac{1}{2}ar(AFC)–\frac{1}{2}\times 2ar(FED)$ 

$4ar(FED)=\frac{1}{2}ar(AFC)$

$ar(FED)=\frac{1}{8}ar(AFC)$