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I am trying to find a specified angle of a triangle.

In triangle $ABC$, $\angle A = 20^\circ$. $D$ and $E$ are points on $AB$ and $AC$, where $AB=AC$. $\angle EBC = 50^\circ$ and $\angle DCB = 60^\circ$. What is the $\angle DEB$ ?

I can show that $\angle DBE = 30^\circ$ and $\angle DCE = 20^\circ$ and I can get all other angles in the interior of the triangle except $\angle DEB$ and $\angle CDE$. Can anyone please help me to solve this? Thanks in advance.

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  • $\begingroup$ Welcome to our site! $\endgroup$ Jan 27, 2015 at 8:58

2 Answers 2

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We can use the property of regular $n$-polygons, that each node sees all other nodes separated by the same angle, being $180^{\circ}/n$.

This makes the regular $18$-polygon a perfect drawing canvas for this problem, in which all angles are a multitude of $10^{\circ}$ (left image): Left: The used polygon property, right: the solution

Let’s also add a smaller $18$-polygon (right image), which:

  • has the same orientation,
  • shares $C$ as a mutual node,
  • touches two of its diagonals, in $A$ and $B$.

Due to similarity, we have that $AB \parallel A’B’$.

Since both $A’$ and $A’’ $, as well as $B’$ and $B’’$ are separated by two edges, we have that $A’B’ \parallel A’’B’’$.

Now using the mentioned polygon property, we see that $\angle A’’B’’B = 30^{\circ}$. $\quad \blacksquare$

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  • $\begingroup$ Your answer is very nice!!! $\endgroup$ Nov 3, 2023 at 23:02
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This is basically Langley's Adventitious Angles problem, a problem that first appeared in a $1922$ Mathematical Gazette, and is well known for being extremely difficult despite its simplistic appearance. Here is a nice piece by piece solution with diagrams.

For the sake of completeness, I will go ahead and add the Wikipedia solution which is attributed to James Mercer.

enter image description here

The idea is to construct a new point $F$ on $AB$ such that $\angle BCF=20^\circ$. After this is done, we can find all of the angles I have listed numbers for by normal angle chasing. Now, by construction we have that $\triangle BCF$, $\triangle BCD$, $\triangle CEF$ as all isosceles, so $|BC|=|CF|=|CD|=|EF|$. Additionally, by construction we have that $\angle FCD=60^\circ$, and since we have two sides the same and an appropriate angle as $60^\circ$, we know that $\triangle CDF$ is in fact equilateral! This gives us the key additional information that $|DF|$ is the same as the other lengths we have described (all outlined in green in the picture). In particular $|DF|=|EF|$, so $\triangle EFD$ is isosceles , so we have the equation $2x+40=180$, which yields the solution $x=70$.

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