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5. 如图8，AB // CD，直线EF分别交AB，CD于点E，F，∠BEF的平分线与∠DFE的平分线相交于点P，求证：∠P = 90°。
答案
5.证明:
∵AB // CD，
∴∠BEF + ∠EFD = 180°.
又
∵EP,FP分别平分∠BEF和∠EFD，
∴∠1 = 1/2∠BEF, ∠2 = 1/2∠EFD.
∴∠1 + ∠2 = 1/2∠BEF + 1/2∠EFD = 1/2(∠BEF + ∠EFD) = 90°.
∴∠P = 180°−(∠1+∠2) = 180°−90° = 90°.
∵AB // CD，
∴∠BEF + ∠EFD = 180°.
又
∵EP,FP分别平分∠BEF和∠EFD，
∴∠1 = 1/2∠BEF, ∠2 = 1/2∠EFD.
∴∠1 + ∠2 = 1/2∠BEF + 1/2∠EFD = 1/2(∠BEF + ∠EFD) = 90°.
∴∠P = 180°−(∠1+∠2) = 180°−90° = 90°.
6.如图9,已知AB // CD,∠C=75°,∠A=25°,求∠E的度数. 
答案
6.解:
∵AB // CD,
∴∠BFE = ∠C = 75°.
∴∠AFE = 180°−∠BFE = 180°−75° = 105°.
∴∠E = 180°−∠A−∠AFE = 180°−25°−105° = 50°.
∵AB // CD,
∴∠BFE = ∠C = 75°.
∴∠AFE = 180°−∠BFE = 180°−75° = 105°.
∴∠E = 180°−∠A−∠AFE = 180°−25°−105° = 50°.
