(1)∠ACB and ∠ADB are the circumferential angles of the corresponding sides of the chord AB, so ∠ACB=∠ADB, and because of de∨BC, ∠ACB=∠AED = ∠ABC and ∠.

(2) Connect OD, then OD⊥DE,DE∥BC, so OD⊥BC, △ABD∽△ADE, then ∠BAD=∠DAC, corresponding to arc BD= arc DC, so OD is the middle vertical line of BC. ∠ ABC = 45, assuming ∠BAC = 90, then △ABC will be an isosceles right triangle, and AO will be the middle vertical line on the side of BC, but actually A, O and D are not * * * lines, so ∠ BAC is not a right angle.

∠ ABC =∠ ADC =∠ ADE = 45, AD⊥AF, △ADF is an isosceles right triangle, and S △ ADF = AD 2/2.

From △ABD∽△ADE, we can get AB/AD = AD/AE, AD 2 = AB * AE, then S△ADF=AB*AE/2.

Since ∠BAC is not a right angle, then the intersection e is the height eg on the side of AB, and EG must be less than AE, then S △ Abe = AB * EG/2 < AB * AE/2. So s △ daf > s △ BAE, good luck! ! ! !