Since ∠BCD is a right angle, it is ∠ 3+∠ 4+∠ 5+∠ 6 = 180. ...
∵∠3=∠4 ∠5=∠6
∴①:2*(∠4+∠5)= 180
∴∠ECF=∠4+∠5=90
∫∠ACD is the outer corner of △ABC.
∴∠acd=∠5+∠6=∠a+∠abc=60+∠ 1+∠2
∵∠5=∠6 ∠ 1=∠2.
∴2∠6=60 +2∠2
Namely: ∠ 6 = 30+∠ 2...②
∫∠6 is the outer corner of △BCF.
∴∠6=∠F+∠2
From ②: f = 30.
In △CEF, ∠ f = 30. ECF = 90 degrees.
∴∠FEC=60
22:
Let ∠ABD=β, ∠ACD=γ.
Because BCD forms a triangle
Then < DBC+< DCB+< BDC = 180.
Then 2β+2γ+A+2/3 ∠ A =180 (1).
Because ∠DBC=2∠ABD, ∠DCB=2∠ACD.
Then ∠ABC=3β, ∠ACB=3γ.
By ∠ ABC+∠ ACB+∠ A = 180.
Then 3β+3γ+∠ A = 180.
Then 2β+2γ+2/3 ∠ A = 120 (2)
Synchronization (1)(2) available
a=60