Does it exist a way to find one side of a triangle in polygon other than using similarity?

Question

The problem is as follows:

Using the figure from below find $EB$. Assume $\triangle ABC$ is isosceles where $AB=AC$ and $\overline{BN} \parallel \overline{AC}$. Find $EB$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&6\,cm\\ 2.&8\,cm\\ 3.&10\,cm\\ 4.&12\,cm\\ 5.&11\,cm\\ \end{array}$

What I attempted to do is sumarized in the figure from below:

As it can be seen I redrawn the figure so it can describe better what it is mentioned in the passage of the problem.

From this part. I relied on triangle similarity.

This would meant:

$\triangle ABE \sim \triangle BED$

Hence:

$\frac{6}{x}=\frac{x}{24}$

Solving this:

$x^2=6\times 24$

Which makes $x=12\,cm$

This later answer does check with the official answer from my book. But since this question is placed before introducing the concept of triangle similarity I am wondering if it could be solved relying using fundamentals of bisector angles and perpendicular bisector segment concepts. Can it be solved only using those?.

As it can be noted, some information is unused which makes me to believe was intentionally put there to use the above mentioned concepts.

All and all, is my solution correct?. I'm not good in this topic thus I need assistance on it. Please, include a drawing in your answer as it could require some construction.

Answer 1

From your diagram, since $\Delta ABN$ is isosceles, $BN = 24$ cm.

We can find $DE$ using Pythaogras, which is $\sqrt{x^2-6^2}$. Similarly, $AE = \sqrt{24^2 - x^2}$.

Now since $\Delta ABE = \Delta NBE$, then we have:

$$\frac{1}{2} BN \cdot DE = \frac{1}{2} BE \cdot AE$$ $$\Rightarrow 24 \sqrt{x^2 - 6^2} = x \sqrt{24^2- x^2}$$ $$\Rightarrow 24^2 (x^2 - 6^2) = x^2 (24^2 - x^2)$$ $$\Rightarrow 24^2 x^2 - 24^2 \cdot 6^2 = 24^2 x^2 = -x^4$$ $$\Rightarrow x^4 = 24^2 \cdot 6^2 \Rightarrow x = ±12.$$

$x$ cannot be $-12$. Now for $x = 12$, the LHS equals the RHS when we plug it in, so $\boxed{x = 12}$.

Answer 2

How about using $\sin \beta$ from two different triangles (ABE and DBE).

Answer 3

Directly or indirectly trigonometry will get in.

Due to the parallelism of sides and equality of forces ( makes for an isosceles triangle) in a parallelogram we can conclude that action of force vectors AB,AC is now a special case ... of a rhombus.

The rhombus diagonals cut at $90^{\circ}...$ convenient trigonometry of a triangle.

$$ BE=\dfrac{6}{\sin \theta}= 24 \sin \theta $$ $$ \sin ^2\theta = \dfrac14 $$ $$ \sin \theta = \dfrac12\to \theta = 30^{\circ}$$

We can verify that $BD$ is half of its half in length computed from a $(30,60,90) $ degrees standard right triangle:

$$ 24 \cos 60^{\circ}\cos 60^{\circ} =6 $$

So there are two equilateral triangles of side $24$ in the diagram shown below:

$BE= 12, $ the answer is Option 4

( Please ignore the circular arcs drawn during construction.)