# How to find the angle in a triangle inside of a quadrilateral?

The problem is as follows:

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

• Why are you posting so many geometry questions? – Parcly Taxel Oct 22 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 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. • How do you find an angle of 40° at the bottom of the figure? – Bernard Oct 22 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 at 14:01
• @Bernard that is wrongly shown. The right part of the angle is $40^0$, not the whole. – Math Lover Oct 22 at 14:02
• @MathLover Note that there is actually a $20-80-80$ isosceles at the very right. – Student1058 Oct 22 at 14:04
• @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 at 14:07