Question

A point

D

is on the side

B C

of an equilateral

△ A B C

, such that

D C = \frac{1}{4} B C

. Prove that

A D^{2} = 13 C D^{2}

.

Accepted Answer

Given, in an equilateral $△ A B C$ , $D C = \frac{1}{4} B C$

To prove, $A D^{2} = 13 C D^{2}$

Construction, construct $A E ⊥ B C$

Since, $△ A B C$ is an equilateral triangle so, $A B = B C = A C$ and $∠ A = ∠ B = ∠ C = 60 °$

Using Pythagoras theorem in a right angle $△ A E C$ , we get
$A C^{2} = A E^{2} + E C^{2}$
$\Rightarrow A E^{2} = A C^{2} - E C^{2} . . . . (i)$

Using Pythagoras theorem in right angle $△ A E D$ , we get
$A D^{2} = A E^{2} + {(E C - C D)}^{2}$
Using identity $(a - b)^{2} = a^{2} + b^{2} - 2 a b$ in ${(E C - C D)}^{2}$ , we get
$A D^{2} = A E^{2} + E C^{2} + C D^{2} - 2 E C . C D . . . . (ii)$

From equation $(i)$ and $(ii)$ , we get

$A D^{2} = A C^{2} - E C^{2} + E C^{2} + C D^{2} - 2 E C . C D$
$\Rightarrow A D^{2} = A C^{2} + C D^{2} - 2 E C . C D$
$\Rightarrow A D^{2} = A C^{2} + {(\frac{1}{4} B C)}^{2} - 2 E C . (\frac{1}{4} B C)$
$\Rightarrow A D^{2} = A C^{2} + \frac{1}{16} B C^{2} - \frac{E C . B C}{2} . . . . (iii)$

In $△ A B E$ and $△ A E C$
$A B = A C$ [side of equilateral triangle]
$A E = A E$ [common angle]
$\Rightarrow △ A B E ≅ △ A E C$ [by RHS congruency of triangle]
$\Rightarrow B E = E C$ [corresponding side of congruent triangle]
$\Rightarrow E C = \frac{B C}{2} . . . . (iv)$ and $A B = B C = A C . . . . (v)$

From equation $(iii), (iv)$ and $(v)$ , we get
$A D^{2} = B C^{2} + \frac{1}{16} B C^{2} - \frac{B C}{2} . \frac{B C}{2}$
$A D^{2} = B C^{2} + \frac{1}{16} B C^{2} - \frac{1}{4} B C^{2}$
$\Rightarrow A D^{2} = \frac{13}{16} B C^{2} . . . . (vi)$
As $D C = \frac{B C}{4} \Rightarrow B C = 4 D C$

Put $B C = 4 D C$ in equation number $(vi)$
$\Rightarrow A D^{2} = \frac{13}{16} {(4 D C)}^{2}$
$\Rightarrow A D^{2} = \frac{13}{16} \times 16 \times D C^{2}$
$\Rightarrow A D^{2} = 13 D C^{2}$

Hence, proved.

H K Dass, Rama Verma and, Bhagwat Swarup Sharma Solutions for Chapter: Similar Triangles, Exercise 7: EXERCISE

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