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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

(iii) $\small ar(ABC)=2ar(BEC)$

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

Answers (1)

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Let's 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 is the midpoint of BC
$\therefore BD = \frac{a}{2}= DE = BE$
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, $\small ar(BDE)=\frac{1}{4}ar(ABC)$ -------(1)

Since $\Delta$ ABC and $\Delta$ BDE are equilateral triangles.

Therefore, $\angle$ ACB = $\angle$ DBE = $60^0$

$\Rightarrow$ BE || AC

$\Delta$ BAE and $\Delta$ BEC are on the same base BE and between the same parallels BE and AC.

Therefore, ar ( $\Delta$ BAE) = ar( $\Delta$ BEC)

$\Rightarrow$ ar( $\Delta$ BAE) = 2 ar( $\Delta$ BED) [since D is the median of $\Delta$ BEC ]

$\Rightarrow$ $\small ar(BDE)=\frac{1}{2}ar(BAE)$

$\Rightarrow$ $\small ar(BDE)=\frac{1}{2}ar(BEC)$

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

Hence proved.

Posted by

manish painkra

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