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In fig, ABC and BDE are two equilateral triangles such that D is the midpoint of BC. If AE intersects BC at F, then the relations
$(i) ar (ADE) = \frac{1}{4} ar (ABC)$

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

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

$(iv) ar (BFE) = ar (AFD)$

$(v) ar (BFE) = 2 ar (FED)$

$(vi) ar (FED) = \frac{1}{8} ar (AFC)$

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(i), (ii) and (iii) are correct

(i), (v) and (iv) are correct

All are correct

None

answer is C.

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Detailed Solution

Given,
Two equilateral triangles are ABE, BDE.
D =midpoint of BC.
Suppose a = side of triangle ABC.
So, its area,

ar (ABC) = \frac{\sqrt{3}}{4} a^{2}

(i) As, D = midpoint of BC. So,

BD = \frac{a}{2}

Here, area of BDE,

ar (BDE) = \frac{\sqrt{3}}{4} {(\frac{a}{2})}^{2} = \frac{\sqrt{3}}{16} a^{2}

\Rightarrow ar (BDE) = \frac{1}{4} ar (ABC)

(ii) Now, DE = median of △BEC Here, each median divides a triangle in two others with the equal area.
Therefore,

ar (BDE) = \frac{1}{2} ar (BEC) \dots (1)

∠ EBC = ∠ BCA = 60^{\circ}

Here, alternate angles in the lines BE and AC with BC =transversal.
So, BE ∥ AC. As,
△BEC and BAE are on same base which is BE so, it lies in the same parallel lines such that BE and AC. Therefore,

ar (BEC) = ar (BAE)

ar (BDE) = \frac{1}{2} ar (BEC) (from (1))

ar (△ BDE) = \frac{1}{2} ar (BAE)

(iii) in the second part ar (BDE) = \frac{1}{2} ar (BEC)

(ED =median of △BEC which divides the triangle in 2 others same area.)
So,

ar (BDE) = \frac{1}{4} ar (ABC)

(from (i))
So,

\frac{1}{4} ar (ABC) = \frac{1}{2} ar (BEC)

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

(iv) ∠ ABD = ∠ BDE = 60^{\circ}

(equilateral triangle angles)
For alternate angles in lines AB and ED where BD =transversal. Therefore,

BA ∥ ED .

Therefore,

ar (BDE) = ar (AED)

(Since, triangles are on same base (ED) and in the same parallel lines AB and line DE)

\begin{matrix} ar (BDE) - ar (FED) = ar (AED) - ar (FED) \end{matrix}

ar (BEF) = ar (AFD)

(v) now in △ ABC,

using Apollonius theorem,

A B^{2} + A C^{2} = 2 A D^{2} + 2 B D^{2}

As,

AB = AC

Put the values,

{AD}^{2} = A B^{2} - B^{2}

A D^{2} = (a)^{2} - {(\frac{a}{2})}^{2}

A D^{2} = a^{2} - \frac{a^{2}}{4} = \frac{3 a^{2}}{4}

AD = \frac{\sqrt{3}}{2} a

Now, in

△ BED,

with the help of Apollonius theorem,
Make a perpendicular from EP on the line BD,
So, EP = median BED
Thus,

{EP}^{2} = {DE}^{2} - {DP}^{2}

{EP}^{2} = {(\frac{a}{2})}^{2} - {(\frac{a}{4})}^{2}

{EP}^{2} = \frac{a^{2}}{4} - \frac{a^{2}}{16} = \frac{3 a^{2}}{16}

\Rightarrow EP = \frac{\sqrt{3} a}{4}

ar (AFD) = \frac{1}{2} \times FD \times AD

ar (AFD) = \frac{1}{2} \times FD \times \frac{\sqrt{3}}{2} a

ar (EFD) = \frac{1}{2} \times FD \times EP

ar (EFD) = \frac{\sqrt{3}}{4} a

Now,

ar (AFD) = \frac{\sqrt{3}}{2} a

ar (EFD) = \frac{\sqrt{3}}{4} a

ar (AFD) = 2 ar (EFD)

Also,

ar (BEF) = ar (AFD)

So,

ar (BFE) = ar (AFD) = 2 ar (EFD)

(vi) In the six part we have ar (BDE) = \frac{1}{4} ar (ABC)

So,

\Rightarrow ar (BEF) + ar (FED) = \frac{1}{4} \times 2 ar (ADC)

2 ar (FED) + ar (FED) = \frac{1}{2} (ar (AFC) - ar (AFD) [Usingpart (v))

3 ar (FED) = \frac{1}{2} ar (AFC) - \frac{1}{2} \times 2 ar (FED)

ar (FED) = \frac{1}{8} ar (AFC)

4 ar (FED) = \frac{1}{2} ar (AFC)

Option 3 is correct.

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