Can you find angle x of the triangle 053

[@antoniopedrofalcaolopesmor6095]

It also could be solved without using auxiliary line ED.
AF =AEcos(x) Then we have
AD = 2AF = 2AEcos(x)
Let's now use Law of sines for triangle ADC :
2AEcosx/sin(DCA)=DC/sinx
2sinxcosxAE=sinDCA*AE
sin(2x) = sin(DCA)
DCA = 2x
Then for triangle ABC we have
30+30+60+x+2x=180
3x=60
x=20