In figure, 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) = \(\frac{1}{4} \) ar (ABC)
(ii) ar (BDE) = \(\frac{1}{2} \) ar (BAE)
(iii) ar (ABC) = 2 ar (BEC)
(iv) ar (BFE) = ar (AFD)
(v) ar (BFE) = 2 ar (FED)
(vi) ar (FED) = \(\frac{1}{8} \) ar (AFC)
[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]

Let us join EC and AD. Draw EP \(\perp \) BC.
Let AB = BC = CA = a, then
BD = \(\frac{a}{2} \) = DE = BE



(i) area of \(\triangle ABC = \frac{\sqrt{3}}{4} × a^2 \)
And area of \(\triangle BDE = \frac{\sqrt{3}}{4} × \frac{a}{2} ^2 \)
= \(\frac{\sqrt{3}}{4} × \frac{a^2}{4} \)
\(\therefore \) ar (BDE) = \(\frac{1}{4} \) ar (ABC)


(ii) Since, \(\triangle \) ABC and \(\triangle \) BED are equilateral triangles.

\(\Rightarrow \) \(\angle \)ACB = \(\angle \)DBE = 60°
\(\Rightarrow \) BE || AC

\(\triangle \)BAE and \(\triangle \)BEC are on the same base BE and between the same parallels BE and AC.

\(\therefore \) ar(\(\triangle \)BAE) = ar(\(\triangle \)BEC)
\(\Rightarrow \) ar(\(\triangle \)BAE) = 2 ar(\(\triangle \)BDE)
[ DE is median of \(\triangle \)EBC. ]
\(\Rightarrow \) ar(\(\triangle \) BDE) = \(\frac{1}{2} \) ar(\(\triangle \)BAE)


(iii) ar(\(\triangle \)ABC) = 4 ar(\(\triangle \)BDE)
[Proved in (i) part]
ar(\(\triangle \)BEC) = 2 ar(\(\triangle \)BDE)
[ \(\because \)DE is median of \(\triangle \) BEC]
\(\Rightarrow \) ar(\(\triangle \)ABC) = 2 ar(\(\triangle \)BEC)


(iv) Since, \(\triangle \)ABC and \(\triangle \)BDE are equilateral triangles.

\(\Rightarrow \) \(\angle \)ABC = \(\angle \)BDE = 60°
\(\Rightarrow \) AB || DE

\(\triangle \)BED and \(\triangle \)AED are on the same base ED and between the same parallels AB and DE.

\(\therefore \) ar(\(\triangle \)BED) = ar(\(\triangle \)AED) Subtracting ar(AEFD) from both sides, we get

\(\Rightarrow \) ar(\(\triangle \)BED) – ar(\(\triangle \)EFD) = ar(\(\triangle \)AED) – ar(\(\triangle \)EFD)
\(\Rightarrow \) ar(\(\triangle \)BEE) = ar(\(\triangle \)AFD)


(v) In right angled \(\triangle \) ABD, we get
\( AD^2 = AB^2 - BD^2 = a^2 - (\frac{a}{2} )^2 = \frac{3a^2}{4} \)
AD = \(\frac{\sqrt{3} a}{2} \)
In right triangle PED
Similarly,
EP = \(\frac{\sqrt{3} a}{4} \)
Area of triangle AFD = (\(\frac{1}{2} \) ) × FD × AD
= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{2} \).....(i) And area of triangle EFD = (\(\frac{1}{2} \)) × FD × EP
= (\(\frac{1}{2} \) ) × FD × \(\frac{\sqrt{3} a}{4} \) .....(ii)

From (1) and (2), we get

ar(\(\triangle \) AFD) = 2 ar((\(\triangle \)EFD)
ar(\(\triangle \)AFD) = ar(\(\triangle \)BEF) [From (iv) part]
\(\Rightarrow \) ar(\(\triangle \)BFE) = 2 ar(\(\triangle \)EFD)


(vi) ar(\(\triangle \)AFC) = ar(\(\triangle \)AFD) + ar(\(\triangle \)ADC)
= ar(\(\triangle \)BFE) + \(\frac{1}{2} \) ar(\(\triangle \)ABC) [From (iv) part]
= ar(\(\triangle \)BFE) + (\(\frac{1}{2} \) ) x 4 x ar(\(\triangle \)BDE) [From (i) part]
= ar(\(\triangle \)BFE) + 2ar(\(\triangle \)BDE)
= 2ar(\(\triangle \)FED) + 2[ar(\(\triangle \)BFE) + ar(\(\triangle \)FED)]
= 2ar(\(\triangle \)FED) + 2[2ar(\(\triangle \)FED) + ar(\(\triangle \)FED)]
[From (v) part]
= 2ar(\(\triangle \)FED) + 2[3ar(\(\triangle \)FED)]
= 2ar(\(\triangle \)FED) + 6ar(\(\triangle \)FED)
= 8ar(\(\triangle \)FED)
\(\Rightarrow \) ar(\(\triangle \)FED) = \(\frac{1}{8} \) ar(\(\triangle \)AFC)