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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.]

Views

Let join the CE and AD and draw $EP \perp BC$. It is given that $\Delta$ABC 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 $\Delta$BDE = $\frac{\sqrt{3}}{4}(\frac{a}{2})^2= \frac{\sqrt{3}}{16}a^2$

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

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