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Though innocent looking - this problem can keep you busy for a very long time and then when you see a solution it seems almost trivial.....
Construct DE so that angle BDE = 60 degrees
Since BDE = 60 deg. is the vertical angle of an isosceles triangle, it follows that triangle BDE is in fact equilateral.
Hence angle ABF = 40 deg. and it follows that triangle ABF is isosceles with BF = FA and as a result FC = FE
By the SAS case of congruency it follows that triangle BFC is congruent to triangle AFE and it follows that AB = AE.
It now follows that ABDE is a kite and AD intersects BE at 90 deg.
And it follows that the elusive angle DAC = 10 deg.