Given ab ∥ CD, EF ∥ CD, ∠ CDE = 150 ° and ∠ bed = 80 °, calculate the degree of ∠ Abe

Given ab ∥ CD, EF ∥ CD, ∠ CDE = 150 ° and ∠ bed = 80 °, calculate the degree of ∠ Abe

Because EF \ \ CD and ∠ CDE = 150 °, so ∠ def = 30 °
And ∠ bed = 80 degree
So ∠ bef = ∠ bed + ∠ def = 30 ° + 80 ° = 110 °
Again ab ∥ CD
So ab ‖ ef
So ∠ Abe = 180 ° - bef = 70 °

As shown in the figure, it is known that AB is parallel to CD, BF bisects ∠ Abe, DF bisects ∠ CDE, ∠ bed = 75 ° to calculate the degree of ∠ BFD

Make a straight line L parallel to AB and CD through point E, then divide ∠ bed into two feet, the upper one is equal to ∠ Abe, and the lower one is always equal to ∠ CDE, because BF bisects ∠ Abe, DF bisects ∠ CDE, and the ∠ FBE + ∠ EDF = 75 ° / 2 = 37.5 ° can be obtained. According to the sum of internal angles of quadrilateral equal to 360 °, it can be known that ∠ BFD = 37.5 °

As shown in the figure, it is known that the bisector of ab ‖ CD, ∠ Abe and ∠ CDE intersects at F, ∠ e = 110 ° and the degree of ∠ BFD is calculated

As shown in the figure, given that the bisectors of ab ‖ CD, ∠ Abe and ∠ CDE intersect at F, ∠ e = 140 °, then the degree of ∠ BFD is______ °．

If eg ∥ AB passes through point E, we can get ∥ Abe + ∥ beg = 180 °, GED + ∥ EDC = 180 ° and ∥ Abe + ∥ CDE + ∥ e = 360 ° and ∥ e = 140 ° and ∥ Abe + ∥ CDE = 220 ° and ∥ FBE + ∥ EDF = 12 (∥ Abe + ∥ CDE) = 110 °. The sum of inner angles of bfde is 360 ° and ∥ BFD = 11