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In Fig. ABC a...

Question

In Fig. $ ABC$ and $ BDE$ are two equilateral triangles such that $ D$ is the mid-point of $ BC$. If $ AE$ intersects $ BC$ at $ F$, show that:

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Solution

Step 1: (i) Prove $a r . (B D E)$ $=$ $\frac{1}{4} a r . (A B C)$ .
Given that
$A B C$ and $B D E$ are two equilateral triangles such that $D$ is the mid-point of $B C$ that is $B D = D C$
Let $A B = x = A C = B C$
Then $B D = \frac{x}{2} = B E = D E$
$a r . (A B C) = \frac{\sqrt{3}}{4} \cdot x^{2} a r . (B D E) = \frac{\sqrt{3}}{4} \cdot {(\frac{x}{2})}^{2} = \frac{\sqrt{3}}{4} \cdot \frac{1}{4} {(x)}^{2} = \frac{1}{4} a r . (A B C)$
Hence, $a r . (B D E)$ $=$ $\frac{1}{4} a r . (A B C)$
Step 2: (ii) Prove $a r . (B D E)$ $=$ $\frac{1}{2} a r . (B A E)$ .
Construction : Join $E C$
$E D$ is median of $∆ B F C$
$a r . (B D E) = \frac{1}{2} a r . (B E C) . . . . . . . . . . . . . . . . . . . . . . (i)$
Due to the altitude interior angles
$\Rightarrow$ $B E ∥ A C$
$\Rightarrow$ $a r . (B E C) = a r . (B E A)$ ……….. $(i i)$ [As both lie on the same base $B E$ and $A C$ between same parallel lines $B E$ and $A C$ ]
From equation $(i)$
$\Rightarrow$ $a r . (B D E)$ $=$ $\frac{1}{2} a r . (B A E)$ .
Step 3: (iii) Prove $a r . (A B C)$ $=$ $2$ $a r . (B E C)$ :
From $(1)$ we get
$\Rightarrow$ $a r . (B D E)$ $=$ $\frac{1}{4} a r . (A B C)$
From $(2)$ we get
$\Rightarrow$ $a r . (B D E)$ $=$ $\frac{1}{2} a r . (B A E)$ .
Using $(1)$ and $(2)$ we get
$\frac{1}{4} a r . (A B C)$ $=$ $\frac{1}{2} a r . (B A E)$
$\Rightarrow a r . (A B C) = 2 a r . (B A E) \Rightarrow a r . (A B C) = 2 a r . (B E C) [∵ a r . (B A E) = (B E C) F r o m (i i)]$
Hence, $a r . (A B C)$ $=$ $2$ $a r . (B E C)$ :
Step 4: (iv) Prove $a r . (B F E)$ $=$ $a r . (A F D)$
Construction: Join $A D$
$∴ A E ∥ E D$
$a r . (E D B) = a r . (E D A)$ [both lie on the same base $E D$ and between same $∥$ lines $E D$ and $A B$ ]
Subtracting $a r . (E D F)$ from both sides
$a r . (E B D) - a r . (E D F) = a r . (E D A) - a r . (E D F) \Rightarrow a r . (B F E) = a r . (A F D)$
Hence $a r . (B F E) = a r . (A F D)$
Step 5: (v) Prove $a r . (B F E)$ $=$ $2 a r . (F E D)$
Let us assume that
$h$ is the height of vertex $E$ , corresponding to the side $B D$ in $∆ B D E$
$H$ is the height of vertex $A$ , corresponding to the side $B C$ in $∆ A B C$ .
While solving Question $(1)$ ,
We observed that
$a r . (B D E)$ $=$ $\frac{1}{4} a r . (A B C)$ .
While solving Question $(4)$
We observed that
$a r . (B F E)$ $=$ $a r . (A F D)$ .
$∴$ $a r . (B F E)$ $=$ $a r . (A F D)$ .
$= 2 a r . (∆ F E D)$
Hence $a r . (B F E)$ $=$ $2 a r . (F E D)$ .
Step 6: (vi) Prove $a r . (F E D)$ $=$ $\frac{1}{8}$ $a r . (A F C)$ .
$a r . (∆ A F C) = a r . (∆ A F D) + a r . (∆ A D C) = 2 a r . (∆ F E D) + \frac{1}{2} a r . (∆ A B C) [Using 5] = 2 a r . (∆ F E D) + \frac{1}{2} (4 a r . (∆ B D E)) [Using result of question 1] = 2 a r . (∆ F E D) + 2 a r . (∆ B D E)$
As $∆ B D E$ and $∆ A E D$ are on the same base and between same parallels
$= 2 a r . (∆ F E D) + 2 a r . (∆ A E D) = 2 a r . (∆ F E D) + 2 (a r . (∆ A F D) + a r . (∆ F E D)) = 4 a r . (∆ F E D) + 4 a r . (∆ F E D) [∵ a r . (∆ A F D) = a r . (∆ F E D) From question 5] = 8 a r . (∆ F E D)$
Hence $a r . (F E D)$ $=$ $\frac{1}{8}$ $a r . (A F C)$ .
Therefore, it is proved that,
$a r . (B D E)$ $=$ $\frac{1}{4} a r . (A B C)$ .
$a r . (B D E)$ $=$ $\frac{1}{2} a r . (B A E)$ .
$a r . (A B C)$ $=$ $2$ $a r . (B E C)$ .
$a r . (B F E)$ $=$ $a r . (A F D)$ .
$a r . (B F E)$ $=$ $2 a r . (F E D)$ .
$a r . (F E D)$ $=$ $\frac{1}{8}$ $a r . (A F C)$ .

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