→ require ∠E, observe the graph, and ∠ABD and ∠ACD can be obtained from the known conditions.

→ Then, according to the properties of the angular bisector, ∠ABE and ∠ECA are not difficult to find.

→∠EMC is the outer corner of △AMB, then ∠EMC=∠A+∠ABM can be found out.

→ In △EMC, find ∠EMC and ∠ECA at the same time, and then find ∠ E.

Solution process

Solution: ∫∠DOC =∠A+∠ABD

∠AOB=∠D+∠ACD

∫∠DOC =∠AOB is here again.

∠A=70，∠D=40

∴∠ABD=40，∠ACD=70

∫BE and CE are equally divided ∠ABD and ∠ACD respectively.

∴∠EBO= 1/2∠ABD=20

∠ECA= 1/2∠ACD=35

∴∠EMC=∠A+∠ABM=90

∴∠e= 180-∠EMC-∠ECA = 180-90-35 = 55

Self-examination/introspection

Draw a picture again on the draft paper according to the conditions given by the topic, so that the combination of numbers and shapes will help to understand the topic and solve the problem faster.