The problem is as follows:

Find the angle $x$ as indicated in the figure from below:

Sketch of the problem

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&36^{\circ}\\ 2.&40^{\circ}\\ 3.&20^{\circ}\\ 4.&30^{\circ}\\ 5.&32^{\circ}\\ \end{array}$

This problem has left me go in circles. I don't know exactly if there's an isosceles or what?. The only thing which I could find was that the angle opposing $50^{\circ}$ is $50^{\circ}$ hence making the upper triangle an isosceles. But that's how far I went. How exactly can this information be used to find the requested angle. Can someone help me with this?.

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    $\begingroup$ Why are you posting so many geometry questions? $\endgroup$ – Parcly Taxel Oct 22 at 12:27
  • $\begingroup$ @ParclyTaxel Hi. I'm sorry for delay. The reason was due I came stuck with these geometry problems long ago. Since I'm recovering from a bronquitis, fever and periodontitis I did not had enough time to publish them before and to attempt resolve them properly. $\endgroup$ – Chris Steinbeck Bell Oct 25 at 23:07
  1. We have $\angle ABC=50^{\circ}$ and $\angle BAC=80^{\circ}$, which implies that $\angle ACB=50^{\circ}$. Therefore $\overline{AB}=\overline{AC}$.
  2. We have $\angle ACE=80^{\circ}$ and $\angle CAE=20^{\circ}$, which implies that $\angle AEC=80^{\circ}$. Therefore $\overline{AC}=\overline{AE}$.
  3. From 1. and 2. we have $\overline{AB}=\overline{AE}$, and note that $\angle BAE=60^{\circ}$. Therefore $\triangle ABE$ is an equilateral triangle. Hence $\angle AEB=60^{\circ}$.
  4. Take a look at $\triangle ADE$. We now have $\angle AEC=80^{\circ}$ and $\angle DAE=40^{\circ}$, which implies that $\angle ADE=40^{\circ}$. Therefore $\overline{AE}=\overline{DE}$. And since $\overline{AE}=\overline{BE}$, we have $\overline{BE}=\overline{DE}$.
  5. $\angle BED=180^{\circ}-\angle AEC-\angle AEB=40^{\circ}$. Therefore $\angle BDE=\frac{180^{\circ}-\angle BED}{2}=70^{\circ}=\angle ADE+x\Longrightarrow \color{red}{x=30^{\circ}}$.

The five blue segments in the image below have the same length, and $\triangle ABE$ is an equilateral triangle.enter image description here

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  • $\begingroup$ How do you find an angle of 40° at the bottom of the figure? $\endgroup$ – Bernard Oct 22 at 14:00
  • $\begingroup$ But why is that so? Two blue segments are true because angles are $40^0$. But how do you know that $3$ blue lines make an equilateral triangle? $\endgroup$ – Math Lover Oct 22 at 14:01
  • $\begingroup$ @Bernard that is wrongly shown. The right part of the angle is $40^0$, not the whole. $\endgroup$ – Math Lover Oct 22 at 14:02
  • $\begingroup$ @MathLover Note that there is actually a $20-80-80$ isosceles at the very right. $\endgroup$ – Student1058 Oct 22 at 14:04
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    $\begingroup$ @Student1058 yes I agree. +1. Please always name the vertices. Life becomes much easier to both raise questions and to answer :) $\endgroup$ – Math Lover Oct 22 at 14:07

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