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.]

Answers (1)


In right angled triangle \DeltaABD, 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 \DeltaPED,
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(\DeltaAFD) = 2. ar(\DeltaEFD) 
Since ar(\DeltaAFD) = ar(\DeltaBEF)

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

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