# ABC is a equilateral triangle. If |BD|=|CF| and |CE|=6, how do you find |AD|=x?

## ABC is a equilateral triangle. If $| B D | = | C F |$ and $| C E | = 6$, how do you find $| A D | = x$?

Sep 12, 2017

$A D = x = 12$ units

#### Explanation:

Draw a line $D G$, parallel to $A C$, as shown in the figure.
$\implies \Delta D B G$ is an equilateral triangle.
Let $D B = y , \implies D G = B G = D B = y$,
Let $\angle D E G = w , \implies \angle F E C = w$,
$\angle E G D = 180 - 60 = {120}^{\circ}$
$\angle E C F = 180 - 60 = {120}^{\circ}$
As $\angle E G D = \angle E C F , \angle D E G = \angle F E C , \mathmr{and} C F = D G$,
$\implies \Delta E G D \mathmr{and} \Delta E C F$ are congruent.
$\implies E G = E C = 6$

Given that $\Delta A B C$ is an equilateral triangle,
$\implies A B = B C$,
$\implies A D + D B = B G + G E + E C$,
$\implies x + y = y + 6 + 6$,
$\implies x = 12$

Hence, $A D = x = 12$ units

Sol. 2)

Draw a line $D H$, parallel to $B C$, as shown in the figure,
$\implies \Delta A D H$ is an equilateral triangle,
$\implies D H = A D = x$,
$\implies H C = D B = y$,
$\implies \Delta D H F \mathmr{and} \Delta E C F$ are similar
$\implies \frac{D H}{H F} = \frac{E C}{C F}$
$\implies \frac{x}{2 y} = \frac{6}{y}$,
$\implies x = 12$ units

Sep 13, 2017

$x = 12.$

#### Explanation:

Let us solve the Problem with the help of Trigonometry.

We want to apply the Sine-Rule to $\Delta D B E , \mathmr{and} \Delta C E F .$

Hence, we need to find, from $\Delta D B E ,$

$\text{sides } D B , B E , \angle D E B , \mathmr{and} , \angle B D E .$

Similarly, in $\Delta C E F ,$ we will require,

$\text{side "CF, } \angle E F C , \mathmr{and} , \angle C E F .$

Suppose that, $B C = a , \mathmr{and} , \angle E F C = \theta .$

$\Delta A B C$ is equilateral.

$\therefore A B = B C = C A = a , \mathmr{and} , \angle B = 60 , \angle E C F = 120.$

In $\Delta C E F , \angle E F C + \angle E C F + \angle C E F = 180.$

$\therefore \theta + 120 + \angle C E F = 180 \Rightarrow \angle C E F = 60 - \theta = \angle D E B .$

Also, BD=AB-AD=a-x, &, as CF=BD=a-x.

Similarly, $B E = B C - E C = a - 6.$

In$\Delta C E F , E C = 6 , C F = a - x , \angle E F C = \theta , \angle C E F = 60 - \theta .$

:. (CE)/(sin/_EFC)=(CF)/(sin/_CEF).......[because," the sine-rule]."

$\Rightarrow \frac{6}{\sin} \theta = \frac{a - x}{\sin} \left(60 - \theta\right) \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right) .$

Next, in DeltaDBC, DB=a-x, BE=a-6, /_DEB=60-theta, &

$\angle B D E = 180 - \left(\angle D B E + \angle D E B\right) = 180 - \left(60 + 60 - \theta\right) = 60 + \theta .$

$\therefore , \text{ by the sine-rule, } \frac{D B}{\sin \angle D E B} = \frac{B E}{\sin \angle B D E} .$

$\Rightarrow \frac{a - x}{\sin} \left(60 - \theta\right) = \frac{a - 6}{\sin} \left(60 + \theta\right) \ldots \ldots \ldots \ldots . . \left(2\right) .$

(1), &, (2) rArr 6/sintheta=(a-6)/sin(60+theta), or,

$\sin \frac{60 + \theta}{\sin} \theta = \frac{a - 6}{6.}$

$\therefore \frac{\sin 60 \cos \theta + \cos 60 \sin \theta}{\sin} \theta = \frac{a}{6} - \frac{6}{6.}$

$\therefore \frac{\sqrt{3}}{2} \cdot \cos \frac{\theta}{\sin} \theta + \frac{1}{2} \cdot \sin \frac{\theta}{\sin} \theta = \frac{a}{6} - 1.$

$\therefore \frac{\sqrt{3}}{2} \cdot \cot \theta + \frac{1}{2} + 1 = \frac{a}{6} , \mathmr{and} , a = 3 \sqrt{3} \cot \theta + 9. \ldots \left(3\right) .$

Now, from $\left(1\right) , \frac{6}{\sin} \theta = \frac{a - x}{\sin} \left(60 - \theta\right) .$

$\therefore \sin \frac{60 - \theta}{\sin} \theta = \frac{a - x}{6.}$

$\therefore \frac{\sin 60 \cos \theta - \cos 60 \sin \theta}{\sin} \theta = \frac{a - x}{6.}$

$\therefore \frac{\sqrt{3}}{2} \cdot \cos \frac{\theta}{\sin} \theta - \frac{1}{2} \cdot \sin \frac{\theta}{\sin} \theta = \frac{a - x}{6.}$

$\therefore 6 \left(\frac{\sqrt{3}}{2} \cdot \cot \theta - \frac{1}{2}\right) = \left(a - x\right) .$

$\therefore 3 \sqrt{3} \cot \theta - 3 - a = - x .$

Finally, by $\left(3\right) ,$ then,

$x = a + 3 - 3 \sqrt{3} \cot \theta = \left(3 \sqrt{3} \cot \theta + 9\right) + 3 - 3 \sqrt{3} \cot \theta .$

$\Rightarrow x = 12 ,$ as Respected CW has already obtained!

Enjoy Maths.!