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# 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)