行云流水

2024-09-25 本文已影响0人无意烦

已知：如图，直线AB//CD, $\measuredangle$ EPM= $\measuredangle$ FQM.求证： $\measuredangle$ AEP= $\measuredangle$ CFQ.

∵ $\measuredangle$ EPM= $\measuredangle$ FQM, ∴ PE//QF,

∴ $\measuredangle PEN=\measuredangle QFN$ ,

∴ $\measuredangle AEN-\measuredangle PEN=\measuredangle CFN-\measuredangle QFN$ ,

即 $\measuredangle AEP=\measuredangle CFQ$ .

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