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.
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.
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 $x=70$.