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In figure, $ \mathrm{ABC} $ and $ \mathrm{BDE} $ are two equilateral triangles such that $ \mathrm{D} $ is the mid-point of $ \mathrm{BC} $. If $ \mathrm{AE} $ intersects $ \mathrm{BC} $ at $ \mathrm{F} $, show that
(i) $ \operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) $
(ii) $ \operatorname{ar}(\mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\mathrm{BAE}) $
(iii) $ \operatorname{ar}(\mathrm{ABC})=2 $ ar $ (\mathrm{BEC}) $
(iv) $ \operatorname{ar}(\mathrm{BFE})=\operatorname{ar}(\mathrm{AFD}) $
(v) $ \operatorname{ar}(\mathrm{BFE})=2 \operatorname{ar}(\mathrm{FED}) $
(vi) $ \operatorname{ar}(\mathrm{FED})=\frac{1}{8} \operatorname{ar}(\mathrm{AFC}) $
[Hint: Join $ \mathrm{EC} $ and $ \mathrm{AD} $. Show that $ \mathrm{BE} \| \mathrm{AC} $ and $ \mathrm{DE} \| \mathrm{AB} $
Given:
\( \mathrm{ABC} \) and \( \mathrm{BDE} \) are two equilateral triangles such that \( \mathrm{D} \) is the mid-point of \( \mathrm{BC} \).
\( \mathrm{AE} \) intersects \( \mathrm{BC} \)
To do:
We have to show that
(i) \( \operatorname{ar}(\mathrm{BDE})=\frac{1}{4} \operatorname{ar}(\mathrm{ABC}) \)
(ii) \( \operatorname{ar}(\mathrm{BDE})=\frac{1}{2} \operatorname{ar}(\mathrm{BAE}) \)
(iii) \( \operatorname{ar}(\mathrm{ABC})=2 \) ar \( (\mathrm{BEC}) \)
(iv) \( \operatorname{ar}(\mathrm{BFE})=\operatorname{ar}(\mathrm{AFD}) \)
(v) \( \operatorname{ar}(\mathrm{BFE})=2 \operatorname{ar}(\mathrm{FED}) \)
(vi) \( \operatorname{ar}(\mathrm{FED})=\frac{1}{8} \operatorname{ar}(\mathrm{AFC}) \)
Solution:
Join \( \mathrm{EC} \) and \( \mathrm{AD} \)
Let $G$ and $H$ be the mid-points of the sides $AB$ and $AC$ respectively.
Join $G$ and $H$.
$GH$ is parallel $BC$.
This implies, by mid point theorem,
$GH=\frac{1}{2}BC$
$GH=BD=DC$
Similarly,
$GD = AH = CH$
$DH = AG = BG$
This implies,
$\triangle ABC$ is divided into four equal equilateral triangles $\triangle BGD, \triangle AGH, \triangle DHC$ and $\triangle GHD$
Therefore,
$\triangle BGD=\frac{1}{4}\triangle ABC$
In $\triangle BDG$ and $\triangle BDE$
$BG=BE$
$BD=BD$
$DG=DE$
Therefore, by SSS congruency,
$\triangle BDG \cong \triangle BDE$
This implies,
$ar(\triangle BDG)=ar(\triangle BDE)$
$ar(\triangle BDE)=\frac{1}{4}ar(\triangle ABC)$
(ii)
$\triangle BDE$ and $\triangle AED$ have common base $DE$ and $DE \| AB$
This implies,
$ar(\triangle BDE) = ar(\triangle AED)$
$ar(\triangle BDE)−ar(\triangle FED) = ar(\triangle AED)−ar (\triangle FED)$
$ar(\triangle BEF) = ar(\triangle AFD)$.......…(i)
$ar(\triangle ABD) = ar(\triangle ABF)+ar(\triangle AFD)$
$ar(\triangle ABD) = ar(\triangle ABF)+ar(\triangle BEF)$ [From (i)]
$ar(\triangle ABD) = ar(\triangle ABE)$........…(ii)
$AD$ is the median of $\triangle ABC$.
This implies,
$ar(\triangle ABD) =\frac{1}{2}ar (\triangle ABC)$
$= 4\times[\frac{1}{2}ar (\triangle BDE)]$
$= 2 ar(ΔBDE)$.......…(iii)
From (ii) and (iii), we get,
$2 ar (\triangle BDE) = ar (\triangle ABE)$
$ar (\triangle BDE) =\frac{1}{2} ar (\triangle BAE)$
(iii) $\triangle ABE$ and $\triangle BEC$ have common base $BE$ and $BE \| AC$
This implies,
$ar(\triangle ABE) = ar(\triangle BEC)$
$ar(\triangle ABF) + ar(\triangle BEF) = ar(\triangle BEC)$
$ar(\triangle ABF) + ar(\triangle AFD) = ar(\triangle BEC)$ [From (i)]
$ar(\triangle ABD) = ar(\triangle BEC)$
$\frac{1}{2}ar(\triangle ABC) = ar(\triangle BEC)$
$ar(\triangle ABC) = 2 ar(\triangle BEC)$
(iv) $\triangle BDE$ and $\triangle AED$ lie on the same base $DE$ and between the parallels $DE$ and $AB$.
This implies,
$ar (\triangle BDE) = ar (\triangle AED)$
Subtracting $ar(\triangle FED)$ from both sides, we get,
$ar (\triangle BDE)−ar (\triangle FED) = ar (\triangle AED)−ar (\triangle FED)$
$ar (\triangle BFE) = ar(\triangle AFD)$
(v) Let $h$ be the height from $E$ to the side $BD$ in $\triangle BDE$ and $H$ the height from $A$ to the side $BC$ in $\triangle ABC$.
$ar (\triangle BDE) = \frac{1}{4}ar (\triangle ABC)$ (Proved)
$ar (\triangle BFE) = ar (\triangle AFD)$ (Proved)
$ar (\triangle BFE) = ar (\triangle AFD)$
$ar(\triangle BFE)= 2 ar (\triangle FED)$
(vi) $ar (\triangle AFC) = ar (\triangle AFD) + ar(\triangle ADC)$
$= 2 ar (\triangle FED) + \frac{1}{2}ar(\triangle ABC)$
$= 2 ar (\triangle FED) +\frac{1}{2}[4ar(\triangle BDE)]$
$= 2 ar (\triangle FED) +2 ar(\triangle BDE)$
$\triangle BDE$ and $\triangle AED$ are on the same base and between the same parallels$
$= 2 ar (\triangle FED) +2 ar (\triangle AED)$
$= 2 ar (\triangle FED) +2 [ar (\triangle AFD) +ar (\triangle FED)]$
$= 2 ar (\triangle FED) +2 ar (\triangle AFD) +2 ar (\triangle FED)$
$= 4 ar (\triangle FED) +4 ar (\triangle FED)$
$ar (\triangle AFC) = 8 ar (\triangle FED)$
$ar (\triangle FED) = \frac{1}{8}ar (\triangle AFC)$
Hence proved.
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