Find value of the angle x Find the value of the angle x.

Plus : Someone could recommend me some good book about this subject ?
 A: Claim
The angle is $20°$, as found in a construction performed with Cinderella.

Proof
If you want to proove this, you could use the following computation using sage. All arithmetic is done using exact algebraic numbers, to avoid floating point rounding issues.
def sin_cos(degrees):
    degrees = QQ(degrees)/360
    z = QQbar.zeta(degrees.denominator())^(degrees.numerator())
    return z.imag(), z.real()
sin10, cos10 = sin_cos(10)
sin20, cos20 = sin_cos(20)
sin30, cos30 = sin_cos(30)
sin40, cos40 = sin_cos(40)
lenAB = 1
lenAC = lenAB*cos20
lenBC = lenAB*sin20
lenAF = lenAC/cos10
lenCF = lenAF*sin10
lenBE = lenBC/cos30
lenCE = lenBE*sin30
lenCF/lenCE == sin40/cos40

The result (printed as True) will show that $\angle FEC=40°$. From the angle sum in $\triangle BCE$ one can conclude $\angle BEC=180°-90°-30°=60°$. So you get $\angle BEF=\angle BEC-\angle FEC=60°-40°=20°$ as the experiment suggested.
A: Purely geometric proof

Let the triangle be $\triangle PQR$ with $\angle QPR = 20^\circ$ and $\angle PRQ = 90^\circ$
Let $O$ be on the side $PQ$ such that $O$ is the centre of a regular 18-gon with radius $OQ$ and one side $AB$ lying on $PR$, where $B$ is between $P$ and $A$, which is possible because that side $AB$ and $OQ$ meet at an angle of $20^\circ$
Then $QR$ clearly passes through another vertex $C$ of the 18-gon
Let $D$ be another vertex of the 18-gon such that $CD \perp OA$
Then $CD$ bisects $OA$ because $\triangle OAC$ is equilateral
Thus $CD$ passes through $P$ because $\triangle OPA$ is isosceles since $\angle AOP = \angle PAO$
Thus $\angle QPC = 10^\circ$ and $\angle CQA = 30^\circ$ and so they are the desired points
Therefore $\angle QAC = 20^\circ$
A: Construction: Construct $\triangle$ PQR $\cong$ $\triangle$ ABC. Construct $\angle$s PST and PSU both equal to 30  intersecting sides PR and PQ at points T and  U respectively. 
Proof: $\angle$ USQ=70. US=UQ. $\triangle$ PTS $\cong$ $\triangle$ PUS. $\triangle$ POT $\cong$ $\triangle$ POU.
$\angle$ POT = $\angle$ POU=90.
Therefore TS=US. $\triangle$ UST is an equilateral triangle and $\angle$s PTO and PUO are both equal to 80. Therefore, $\angle$s UTQ and UQT are both equal to 40.
Therefore, $\angle$ QTS=20 and $\angle$ RQT=30
Now $\triangle$ BCE $\cong$ QRT and $\triangle$ DCE $\cong$ SRT. 
That proves x=20.

