As shown in the figure, CD ‖ EF, ∠ C ＋ f = ∠ ABC, verification: ab ‖ GF

As shown in the figure, CD ‖ EF, ∠ C ＋ f = ∠ ABC, verification: ab ‖ GF

It is proved that: let CB's extension line intersect Fe's extension line at h, FG at K. ∵ CD ∥ Fe (known) ∵ f = ∵ FHK (two lines are parallel and the same angle is equal), ∵ HKG = ∵ C + ∵ FHK (the sum of the external angles of a triangle is equal to the sum of two non adjacent internal angles) ∵ HKG =

As shown in the figure, CD ∥ AB, ∠ DCB = 70?, ∠ CBF = 20?, ∠ EFB = 130?, what is the position relationship between the straight line EF and ab? Why?

It is proved that: ∵ CD ‖ AB, ∵ ABC = ∠ DCB = 70 °; also ∵ CBF = 20 °, ∵ ABF = ∠ ABC - ∠ CBF = 70 ° - 20 ° = 50 °; ∵ ABF + ∠ EFB = 50 ° + 130 ° = 180 °; ∵ EF ‖ AB (complementary inner angle on the same side, two lines parallel)