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