Question

如图,AD是三角形ABC的角平分线,点D为AD上一点DM平行AC交于M,PN平行AB交于AC于N,求证PA平方角MPA

Answer 1

结论应为：PA平分∠MPN.

证明,∵AD平分∠BAC,∴∠DAB=∠DAC,

∵PM∥AC,∴∠APM=∠DAC,

∵PN∥AB,∴∠APN=∠DAB,

∴∠APM=∠APN,

∴AP平分∠MPN.