Another angle inside a pentagon I saw this problem recently:
An angle inside a regular pentagon
My question is what would be the geometric constructions to find the $\theta$ angle if we draw $42°$ from the vertices of a regular pentagon.
I saw @timon92's answer in the link above, but it don't work for this particular case. I know the answer is $60°$, however my attempts to solve it are reduced to draw it backwards. I mean, to suppose it's $60°$ and then verify that the angles are $42°$.

 A: This is a different proof using Morley's Trisector Theorem.  This theorem is usually proven using trigonometry.  However, a geometric proof of this theorem can be found here (this is a backward proof) and there (this is a forward proof).

Let $O$ be the circumcenter of the pentagon $ABCD$.  Each side of the pentagon subtends $\dfrac{360^\circ}{5}=72^\circ$ at $O$.  Therefore, $$\angle COE=2\cdot 72^\circ=144^\circ=3\cdot 48^\circ\,.$$  Since $COE$ is an isosceles triangle, we have
$$\angle OCE=\angle OEC=18^\circ\,.$$
As $CDE$ is an isoscels triangle with $\angle CDE=108^\circ$, we conclude that $$\angle DCE=\angle DEC=36^\circ\,.$$  Because $\angle FCD=\angle FED=42^\circ$, we obtain
$$\angle FCE=\angle FEC=6^\circ\,.$$
Therefore, $FC$ and $FE$ are angular trisectors of the triangle $COE$.
Let the angular bisector of $\angle FCO$ meet the circumcircle of the triangle $BOC$ at $P$.  Then, $\angle OCP=6^\circ$, so that $$\angle OBP=\angle OCP=6^\circ\,.$$
Because $\angle BOC=72^\circ$ and $BOC$ is an isosceles triangle, we have
$$\angle OCB=\angle OBC=54^\circ\,.$$
Ergo,
$$\angle COP=\angle PBC=\angle OBC-\angle OBP=54^\circ-6^\circ=48^\circ\,.$$
By symmetry, if $Q$ is the image of reflection of $P$ with respect to the line $OD$, then $$\angle EOQ=48^\circ\,.$$  Therefore, $OP$ and $OQ$ trisect $\angle COE$.  Moreover, $CF$ and $CP$ trisect $\angle OCE$, and $EF$ and $EQ$ trisect $\angle OEC$.  By Morley's Trisector Theorem, $FPQ$ is an equilateral triangle.
By symmetry, $OP=OQ$.  As $\angle POQ=48^\circ$, we get
$$\angle OPQ=\angle OQP=66^\circ\,.$$
Because 
$$\angle OPB=\angle OCB=54^\circ\,,$$
we get
$$\angle FPQ+\angle OPQ+\angle OPB=60^\circ+66^\circ+54^\circ=180^\circ\,.$$
Therefore, $P$ lies on $FB$.  Similarly, $Q$ lies on $FA$.  Thus,
$$\angle AFB=\angle PFQ=60^\circ\,.$$

If you do not wish to invoke Morley's Trisector Theorem, you can still prove that $FPQ$ is an equilateral triangle.  Let $X$ be the intersection of the lines $CP$ and $EQ$, $Y$ the intersection of $EF$ and $OP$, and $Z$ the intersection of $CF$ and $OQ$.  Clearly, $PZ$ and $QY$ intersects at a point $U$ on the line $FX$.

It is easily seen that
$$\angle YFZ=\angle CFE=180^\circ-6^\circ-6^\circ=168^\circ\,,$$ so that
$$\angle UFY=\angle UFZ=84^\circ\,.$$
Likewise,
$$\angle QXP=\angle EXC=180^\circ -2\cdot(6^\circ+6^\circ)=156^\circ\,,$$
so
$$\angle PXU=\angle QXU=78^\circ\,.$$
Since $\angle POQ=48^\circ$, so
$$\angle XOP=\angle XOQ=24^\circ\,.$$
We also have
$$\angle OQX=\angle OPX=\angle PCO+\angle COP=6^\circ+48^\circ=54^\circ\,.$$
Next, observe that
$$\angle QZF=\angle OZC=180^\circ-2\cdot 48^\circ-2\cdot 6^\circ=72^\circ\,.$$
Since $P$ is the intersection of the internal angular bisectors of the triangle $ZOC$, we get
$$\angle QZU=\angle UZF=36^\circ\,.$$
Similarly, $$\angle PYU=\angle UTD=36^\circ\,.$$
Ths shows that
$$\angle ZPO=180^\circ-\angle POZ-\angle PZO=96^\circ\,,$$
whence
$$\angle XPU=\angle XPZ=\angle ZPO-\angle OPX=96^\circ-54^\circ=42^\circ\,.$$
This shows that
$$\angle XUP=180^\circ-\angle  PXU-\angle XPU=60^\circ\,.$$
Similarly,
$$\angle XUQ=60^\circ\,.$$
However,
$$\angle FUY=\angle OYU+\angle UOY=24^\circ+36^\circ=60^\circ\,.$$
This shows that $$\angle YUP=180^\circ-\angle FUY-\angle XUP=60^\circ$$
too.  In conclusion,
$$\angle FUY=\angle YUP=\angle PUX=60^\circ\,.$$ 
By symmetry,
$$\angle FUZ=\angle ZUQ=\angle QUX=60^\circ\,.$$
By symmetry, it is clear that $UP=UQ$.  Now, observe that the triangles $FUY$ and $PUY$ have $$\angle FUY=\angle PUY\,,\,\,UY=UY\,,\,\,\text{ and }\angle UYF=\angle UYP\,.$$  Therefore, they are congruent triangles.  This means $UF=UP$.  That is,
$$UF=UP=UQ\,.$$
This means $U$ is the circumcenter of the triangle $FPQ$.  Since $$\angle PUQ=\angle QUF=\angle FUP=120^\circ\,,$$
we deduce that
$$\angle PFQ=\angle QPF=\angle FQP=60^\circ\,.$$
Therefore, $FPQ$ is an equilateral triangle.
A: Let the edge length be $1$
Law of sines gives us: $$|FC|=\frac{\sin 54°}{\sin 84°}$$
Law of cosines gives us $$|FB|^2=1+\left(\frac{\sin 54°}{\sin 84°}\right)^2-2\left(\frac{\sin 54°}{\sin 84°}\right)\cos 66°=1$$
So $|FA|=|FB|=1$ and triangle $AFB$ is equilateral.
A: This is actually pretty easy if you do it backwards. To make this idea work you have to construct a point $F'$ satisfying the thesis and prove that it coincides with $F$.
Let $F'$ be such that $ABF'$ is equilateral. Then $\angle F'AE= 108^\circ -60^\circ=48^\circ$ so since $AE=AF'$ we have $\angle AEF' = 90^\circ - \frac 12\angle F'AE = 66^\circ$. Hence $\angle F'ED = 108^\circ - 66^\circ = 42^\circ$. We prove analogously that $\angle DCF'=42^\circ$. Hence $F'$ coincides with $F$ and consequently $\angle AFB = \angle AF'B = 60^\circ$.
A: If the pentagon is regular, then we know the the angles FCB, FEA (the interior angles of a pentagon-42). Given that we know the angle EFC (360-sum of other interior angles), how can we figure out the value of theta (hint: what is the sum of all angles around a point?). Lemme know if you need some more elaboration
