Take a pair of triangle ruler and splice it according to the way shown in Fig. 1, fix the triangle ruler ADC, and rotate the triangle ruler ABC clockwise around point a to get the triangle ruler Let's ask when α is how many degrees, can make ab ‖ DC? 2 rotate to the position of ③, then how many degrees is α?

Take a pair of triangle ruler and splice it according to the way shown in Fig. 1, fix the triangle ruler ADC, and rotate the triangle ruler ABC clockwise around point a to get the triangle ruler Let's ask when α is how many degrees, can make ab ‖ DC? 2 rotate to the position of ③, then how many degrees is α?

Take a pair of triangle plates and splice them according to figure 1, fix the triangle plate ADC, and rotate the triangle plate ABC clockwise around point a for an angle of α (0 ° < α ≤ 45 °) to get △ ABC ', as shown in the figure
Let's ask: (1) when α is the degree, it can make ab ‖ DC in Figure 2;
(2) Connect BD, when 0 °＜ α≤ 45 °, explore the size change of ∠ DBC ′ + ∠ CAC ′ + ∠ BDC value, and give your proof. Test points: the nature of rotation; the determination of parallel lines; triangle inner angle and theorem. Special topic: inquiry type. Analysis: (1) to make ab ‖ DC, as long as ∠ CAC ′ = 15 ° can be proved
(2) When 0 °＜ α≤ 45 °, there is always △ EFC ′. According to ∠ EFC ′ = ∠ BDC + ∠ DBC ′, and because ∠ EFC ′ + ∠ FEC ′ + ∠ C ′ = 180 °, we get ∠ BDC + ∠ DBC ′ + ∠ C + α + ∠ C ′ = 180 °, then ∠ DBC ′ + ∠ CAC ′ + ∠ BDC = 105 °. (1) from the meaning ∠ CAC ′ = α,
In order to make ab ‖ DC, it is necessary to ∠ BAC = ∠ ACD,
∴∠BAC=30°,α=∠CAC′=∠BAC′-∠BAC=45°-30°=15°,
That is to say, when α = 15 ° ab ‖ DC
(2) There is no change in the value of connecting BD, ∠ DBC ′ + ∠ CAC ′ + ∠ BDC, which is always 105 degrees,
There is always △ EFC ′ when 0 ° < α≤ 45 ℃
∵∠EFC′=∠BDC+∠DBC′,∠CAC′=α,∠FEC′=∠C+α,
And ∵ EFC ′ + ∵ FEC ′ + ∵ C ′ = 180 °,
∴∠BDC+∠DBC′+∠C+α+∠C′=180°,
And ∵ C ′ = 45 °, C = 30 °,
∴∠DBC′+∠CAC′+∠BDC=105°．