As shown in the figure, given that the bisectors of ab ‖ CD, ∠ Abe and ∠ CDE intersect at F, ∠ e = 140 & ordm;, find the degree of ∠ BFD?

As shown in the figure, given that the bisectors of ab ‖ CD, ∠ Abe and ∠ CDE intersect at F, ∠ e = 140 & ordm;, find the degree of ∠ BFD?

If FG is parallel to ab through F, the internal stagger angle of AB is equal, so ∠ ABF = ∠ BFG ∠ FDC = ∠ DFG, so ∠ BFD = ∠ ABF + ∠ CDF = ∠ Abe / 2 + ∠ CDE / 2, and EH is parallel to ab through e, then AB is complementary to the internal angle of AB, so ∠ Abe + ∠ beh = 180, ∠ CDE + ∠ CDH = 180, add ∠ e + ∠ Abe + ∠ CDE = 360 ∠ Abe + ∠ CDE = 360-140 = 220, so

As shown in the figure, ab ∥ CD, BF bisection ﹥ Abe, DF bisection ﹥ CDE, ﹥ BFD = 55 ° to calculate the degree of ﹥ bed

As shown in the figure, the crossing points E and F are respectively eg ‖ AB, FH ‖ ab. ∵ eg ‖ AB, FH ‖ AB, ∵ 5 = ∠ Abe, ∵ 3 = ∠ 1; and ∵ ab ‖ CD, ∵ eg ‖ CD, FH ‖ CD, ∵ 6 = ∠ CDE, ∵ 4 = ∵ 2, ? 1 + ∵ 2 = ? 3 + ? 4 = ? BFD = 55 °.∵ BF bisection ∠ Abe, DF bisection ∠ CDE, ∵ Abe = 2

As shown in the figure, ab ∥ CD, ∠ bed = 110 °, BF bisection ∥ Abe, DF bisection ∥ CDE, then ∥ BFD = ()
A. 110°B. 115°C. 125°D. 130°

EM ∥ AB, FN ∥ AB, ∥ ab ∥ CD, ∥ EM ∥ ab ∥ CD ∥ FN, ∥ Abe + ∥ BEM = 180 °, ∥ CDE + ∥ DEM = 180 °, ∥ Abe + ∥ bed + ∥ CDE = 360 °, ∥ bed = 110 °, ∥ Abe + ∥ CDE = 250 °, ∥ BF ∥ Abe, DF ∥ CDE, ∥ ABF = 12