The last question in the mid-term evaluation test paper of the seventh grade mathematics teaching in Chongqing No.1 Middle School (Part Two)
By ∠ABF=∠AFB, ∠AFC=∠ACF:
AB=AF=AC
So ∠ABD=∠ACD
Then ∠1+∠ Abd = ∠ ADC = ∠ AFC+∠ 2 = ∠ ACF+∠ 2 = ∠ ACD+2 ∠ 2.
Therefore ∠ 1=2∠2.
Similarly < 3 = 2 < 4.