这是一道数学题，我怎么理解？---**题目** 27. (10分) 如图1,点 E 是直线 AB 上一点,F 是直线 CD 上一点,AB∥CD. (1)求证: ∠P = ∠PEA + ∠PFC; (2)如图2,∠PFC = ∠PFQ, FQ 与∠AEP 的平分线交于点 Q, 与 PE 相交于点 M, 若∠EMF = 120°, 求∠P + ∠Q 的度数; (3)如图3, EQ 平分∠AEP, FM 平分∠PFD, FN∥EQ, 当∠P 的大小不变时, 下列结论: ①∠PFC + ∠NFD 的度数不变; ②∠MFN 的度数不变, 其中有且只有一个是正确的, 请你写出正确的结论并说明理由. **Figures Description:** * **图 1 (Figure 1):** * Two parallel lines AB and CD. Line AB contains point E, line CD contains point F. * Point P is in the region between the two parallel lines. * Line segments PE and PF connect point P to points E and F respectively. * Angles involved: ∠PEA, ∠PFC, ∠EPF (labeled as ∠P). * **图 2 (Figure 2):** * Two parallel lines AB and CD. Line AB contains point E, line CD contains point F. * Point P is in the region between the two parallel lines. * Line segments PE and PF connect point P to points E and F respectively. * A ray FQ originates from F. ∠PFC = ∠PFQ is indicated by the text. * A ray EQ originates from E, which is the bisector of ∠AEP. * FQ intersects EQ at point Q. * FQ intersects PE at point M. * Angles involved: ∠PEA, ∠PFC, ∠P, ∠Q, ∠EMF. * **图 3 (Figure 3):** * Two parallel lines AB and CD. Line AB contains point E, line CD contains point F. * Point P is in the region between the two parallel lines. * Line segments PE and PF connect point P to points E and F respectively. * Ray EQ originates from E and bisects ∠AEP. * Ray FM originates from F and bisects ∠PFD. * Ray FN originates from F. FN is parallel to EQ. * Angles involved: ∠AEP, ∠PFD, ∠PFC, ∠NFD, ∠MFN.