Question

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

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

(iii) $ar\left(ABC\right)=2ar\left(BEC\right)$

(iv) $ar\left(BFE\right)=ar\left(AFD\right)$

(v) $ar\left(BFE\right)=2ar\left(FED\right)$

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

Answer 1

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

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

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

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

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

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

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

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

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

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

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

(iii) $ar\left(BDE\right)=\frac{1}{2}ar\left(BEC\right)$

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

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

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

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

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

Subtracting $ar\left(FED\right)$ from both sides,

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

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

(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 $BED$

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

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

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

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

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

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

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

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

Now,

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

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

$ar\left(AFD\right)=2ar\left(EFD\right)$

Also, $ar\left(BEF\right)=ar\left(AFD\right)$

So, $ar\left(BFE\right)=ar\left(AFD\right)=2ar\left(EFD\right)$

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

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

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

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

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

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