As shown in the figure, given the angle Abe + angle CDE + angle bed = 360 ° to prove ab ‖ CD

As shown in the figure, given the angle Abe + angle CDE + angle bed = 360 ° to prove ab ‖ CD

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As shown in the figure, given the straight line ab ‖ CD, when the point E is between the straight line AB and CD, ∠ bed = ∠ Abe + ∠ CDE holds; when the point E is outside the straight line AB and CD, the following relation holds ()
A. Either ∠ bed = ∠ Abe + ∠ CDE or ∠ bed = ∠ Abe - ∠ cdeb.} bed = ∠ Abe - ∠ CDEC.} bed = ∠ CDE - ∠ Abe or ∠ bed = ∠ Abe - ∠ CDed.} bed = ∠ CDE - ∠ Abe

As shown in the figure, when e is above AB, make ef ∥ AB, ∫ CD ∥ AB, ∥ EF ∥ CD, ∥ fed = ∥ 3, ∥ 1 = ∥ 2 through e, so ∥ bed = ∥ fed - ∥ Feb = ∥ CDE - ∥ Abe; when e is below DC, similarly, ∥ bed = ∥ Abe - ∥ CDE can be obtained

As shown in the figure, if AB is parallel to CD, BF bisector angle Abe, DF bisector angle CDE, and angle bed = 75 degrees, then angle BFD =?

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