# Q : 5        In Fig.$\small 9.33$, 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) $\small ar(BFE)=2ar(FED)$[Hint : Join EC and AD. Show that  $\small BE\parallel AC$ and  $\small DE\parallel AB$, etc.]

M manish painkra

In right angled triangle $\Delta$ABD, we get

$\\AB^2 = AD^2+BD^2\\ AD^2 = AB^2-BD^2$
$=a^2 - \frac{a^2}{4}=\frac{3a^2}{4}$
$AD=\frac{\sqrt{3}a}{2}$

So, in $\Delta$PED,
$PE^2 = DE^2-DP^2$
$\\=(\frac{a}{2})^2-(\frac{a}{4})^2\\ =\frac{3a^2}{16}$
So,  $PE=\frac{\sqrt{3}a}{4}$

Therefore, the Area of $\Delta AFD =1/2 (FD)\frac{\sqrt{3}a}{2}$..........(i)

And, Area of triangle $\Delta EFD =1/2 (FD)\frac{\sqrt{3}a}{4}$...........(ii)

From eq (i) and eq (ii), we get
ar($\Delta$AFD) = 2. ar($\Delta$EFD)
Since ar($\Delta$AFD) = ar($\Delta$BEF)

$\Rightarrow$$\small ar(BFE)=2ar(FED)$

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