In ▱ ABCD, the bisector BC of ∠ bad is at point E, and the bisector DC is at point F (1) Prove CE = CF in Figure 1; (2) if ∠ ABC = 90 ° and G is the midpoint of EF (as shown in Figure 2), write the degree of ∠ BDG directly; (3) if ∠ ABC = 120 ° and FG ‖ CE, FG = CE, connect dB and DG respectively (as shown in Figure 3), and calculate the degree of ∠ BDG

In ▱ ABCD, the bisector BC of ∠ bad is at point E, and the bisector DC is at point F (1) Prove CE = CF in Figure 1; (2) if ∠ ABC = 90 ° and G is the midpoint of EF (as shown in Figure 2), write the degree of ∠ BDG directly; (3) if ∠ ABC = 120 ° and FG ‖ CE, FG = CE, connect dB and DG respectively (as shown in Figure 3), and calculate the degree of ∠ BDG

(1) It is proved that: as shown in Figure 1, ∵ AF bisects ∵ bad, ∵ BAF = ∵ DAF, ∵ quadrilateral ABCD is parallelogram, ∵ ad ∥ BC, ab ∥ CD, ∵ DAF = ∵ CEF, ∵ BAF = ∵ f. ∵ CE = cf. (2) connecting GC, BG, ? quadrilateral ABCD is parallelogram, ∵ ABC = 90 °, ∵ quadrilateral ABCD is rectangle, ? AF bisects ∵ bad, ∵ DAF = ∵ BAF = 45 °, ? DCB = 90 °, ? AF bisects ∫ DAF = ∫ DAF = 45, The results show that: AB, the \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\= 90 ° and ∵ D (3) extending the intersection of AB and FG to h, connecting HD. ∵ ad ∥ GF, ab ∥ DF, ∵ quadrilateral ahfd is parallelogram ∵ ABC = 120 °, AF bisection ∵ bad ∥ DAF = 30 °, ADC = 120 °, DFA = 30 ° DH, △ DHF are congruent equilateral triangles  DH = DF,  Bhd =  GFD = 60 °  FG = CE, CE = CF, CF = BH,  BH = GF. In  Bhd and  GFD,  DH = DF,  Bhd =  GFD = GF,  BDH =  GDF  BDG =  BDH +  HDG =  GDF +  HDG = 60 °

In the parallelogram ABCD, the bisector of the angle bad intersects the line BC and the point E, and the intersecting line DC and the point F
If the angle ABC equals 90 degrees and G is the midpoint of EF, find the degree of angle BDG
If the angle ABC is equal to 120 degrees, FG / / CE, connect dB and DG respectively, and find the degree of the angle BDG

2. Yes, it means FG / / CE. It seems that you are very good at mathematics. I'm now in the third grade of junior high school. I'm in the top five in several books, but I'm not very good at mathematics. I'm weak in the top ten. How can I learn mathematics well? Usually the teacher talks about it, that is, when I do a problem, I don't know what knowledge I should have. Help me.

In the parallelogram ABCD, the bisector of the angle bad intersects the straight line BC at point E and the straight line DC at point F
2) If ∠ ABC = 90 ° and G is the midpoint of EF (as shown in Figure 2), write the degree of ∠ BDG directly
Figure not provided
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It is proved that: ∵ ABCD is a parallelogram, and ∵ ABC = 90 °, the parallelogram ABCD is a rectangle, ∵ bad = eclf = 90 ° and ab = DC. ∵ bad = 90 ° and ∵ BAE = bad / 2, ∵ BAE = 45 ° and ∵ Abe = 90 ° and ∵ ab = be