TITLE: Right angled triangle inscribed in a regular pentagon; find angle
QUESTION [5 upvotes]: $ABCDE~$ is a regular pentagon. Find $x$.

I am having a lot of trouble with this geometry problem. Simply using trigonometry seems to always lead me to a dead end and I can't leverage any circle theorems (I chose the midpoint of the long side as the radius as it is a right triangle, but it didn't lead me anywhere). I noticed that the angles of the triangle are akin to dividing the triangle you get when you divide the pentagon into $10$ pieces but I also do not know how to use that information. I tried cutting up the pentagon and gluing it back together, combining with different pentagons. I don't know any more ways of solving geometry problems any more.

REPLY [2 votes]: The first thing I would acknowledge that it is a bit of a convoluted proof and is not that straightforward to come to the conclusion without drawing and getting some insights. We know that perpendicular from vertex $C$ to diagonal $BE$ and to side $AE$ will make angle of $36^0 (\angle ICK = 36^0)$ and $\angle BCI = 18^0$. That will also mean $\angle DCI = 90^0$.
Now we are given $\angle DKI = 90^0, \angle KDI = 36^0, \angle DIK = 54^0$. This shows we are talking about the unique construct shown in the diagram.
That means $x = 18^0$.
Now a few interesting facts related to this problem.
We draw perp from two adjacent vertices $C, D$ to the opposite diagonal $BE$ and let the perp lines intersect sides $AE$ and $AB$ at point $H$ and $I$ respectively.
Though it is obvious based on angles but to call out points $H$ and $I$ are not the midpoints of sides $AE$ and $AB$.
Now it is easy to show $\angle EDH = \angle BCI = 18^0$, and CDHI is a rectangle with its center at $J$.
Then we draw the circumscribing circle to this rectangle. It intersects the side $AE$ of the pentagon at point $H$ but also at the midpoint $K$. Similarly points $F$ and $I$ on the side AB. It is $\triangle DKI$ that the question is referring to.