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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?.

Chris Steinbeck Bell
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    Why are you posting so many geometry questions? – Parcly Taxel Oct 22 '20 at 12:27
  • @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. – Chris Steinbeck Bell Oct 25 '20 at 23:07

1 Answers1

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  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

  • How do you find an angle of 40° at the bottom of the figure? – Bernard Oct 22 '20 at 14:00
  • 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? – Math Lover Oct 22 '20 at 14:01
  • @Bernard that is wrongly shown. The right part of the angle is $40^0$, not the whole. – Math Lover Oct 22 '20 at 14:02
  • @MathLover Note that there is actually a $20-80-80$ isosceles at the very right. –  Oct 22 '20 at 14:04
  • @MathLover Sorry for misunderstanding. I am new to Geogebra. Yes $40^{\circ}$ is the right part of the angle. And the whole angle is $70^{\circ}$. –  Oct 22 '20 at 14:05
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    @Student1058 yes I agree. +1. Please always name the vertices. Life becomes much easier to both raise questions and to answer :) – Math Lover Oct 22 '20 at 14:07
  • I see. I am very sorry. Thanks for the advice. I will keep it in mind. –  Oct 22 '20 at 14:09
  • @Student1058 I wonder how did you came to the conclusion that the blue lines make an isosceles? And I'm not referring to the $20-80-80$. As indicated I think it would help a lot if you include the names for the vertex in this figure. After long inspection I noted that you might had used the identity that the sum of the two interior angles in a triangle equals to the exterior of the third. But how did you come to the conclusion that the whole angle is $70^{\circ}$? Can you include this explanation please? – Chris Steinbeck Bell Oct 25 '20 at 23:20
  • @Student1058 Howdy!. Don't be worry about using geogebra, it takes a bit of patience to do it. As for me, I prefer using Inkscape since it seems easier to adjust the figures the way I need them. Although it isn't exactly accurate when it comes to put the angles in the right proportion as indicated in your solution which as noted above requires the explanation of how did you come to the conclusion that such side is $70^{\circ}$. – Chris Steinbeck Bell Oct 25 '20 at 23:22
  • @ChrisSteinbeckBell I completely edited my answer and added full explanation. Hope this helps. –  Oct 26 '20 at 02:49