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In Fig.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 (i) ar (BDE) = 1/4 ar (ABC)

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

(i)        \small ar(BDE)=\frac{1}{4}ar(ABC)

[Hint: Join EC and AD. Show that \small BE\parallel AC and   \small DE\parallel AB , etc.]

        

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Let join the CE and AD and draw EP \perp BC. It is given that \DeltaABC and \Delta BDE is an equilateral triangle.
So,   AB =BC = CA = a  and D i sthe midpoint of BC 
therefore, BD = \frac{a}{2}= DE = BE
(i) Area of \Delta ABC = \frac{\sqrt{3}}{4}a^2 and 
Area of \DeltaBDE = \frac{\sqrt{3}}{4}(\frac{a}{2})^2= \frac{\sqrt{3}}{16}a^2

Hence, \small ar(BDE)=\frac{1}{4}ar(ABC)

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