If PE |||| L1is made at the intersection p, then PE ||| L2 (because L 1||L2) is ∠ 1=∠CPE, ∠2=∠DPE, ∠.
So ∠3=∠ 1+∠2.
When point P coincides with point A, ∠ 1 = 0 ∠ 2 = ∠ 3 (two straight lines are parallel with equal internal angles). At this time ∠3=∠ 1+∠2 still holds.
When point P and point B coincide, ∠ 2 = 0 ∠ 1 = ∠ 3 (two straight lines are parallel with equal internal angles). At this time ∠3=∠ 1+∠2 still holds.
∠ 1, ∠2 and ∠3 satisfy ∠3=∠ 1+∠2.
2. From the proof process of the problem 1, it can be seen that as long as the point P moves on the line segment AB, the relationship between ∠ 1, ∠2 and ∠3 has not changed.
3. When P moves out of A,
If PE |||| L1is made at the intersection p, then PE ||| L2 (because L 1||L2) is ∠ 1=∠CPE, ∠2=∠DPE, ∠.
When p moves out of b,
If PE |||| L1is made at the intersection p, then PE ||| L2 (because L 1||L2) is ∠ 1=∠CPE, ∠2=∠DPE, ∠.