如图,BD是∠ABC的平分线,BA=BC,点P在BD的延长线,PM⊥AD,PN⊥CD,点M\N分为垂足,求证:PM=PN.
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![如图,BD是∠ABC的平分线,BA=BC,点P在BD的延长线,PM⊥AD,PN⊥CD,点M\N分为垂足,求证:PM=PN.](/uploads/image/z/11315497-49-7.jpg?t=%E5%A6%82%E5%9B%BE%2CBD%E6%98%AF%E2%88%A0ABC%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%2CBA%3DBC%2C%E7%82%B9P%E5%9C%A8BD%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%2CPM%E2%8A%A5AD%2CPN%E2%8A%A5CD%2C%E7%82%B9M%5CN%E5%88%86%E4%B8%BA%E5%9E%82%E8%B6%B3%2C%E6%B1%82%E8%AF%81%EF%BC%9APM%3DPN.)
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如图,BD是∠ABC的平分线,BA=BC,点P在BD的延长线,PM⊥AD,PN⊥CD,点M\N分为垂足,求证:PM=PN.
如图,BD是∠ABC的平分线,BA=BC,点P在BD的延长线,PM⊥AD,PN⊥CD,点M\N分为垂足,求证:PM=PN.
如图,BD是∠ABC的平分线,BA=BC,点P在BD的延长线,PM⊥AD,PN⊥CD,点M\N分为垂足,求证:PM=PN.
证明:
∵BD平分∠ABC
∴∠ABD=∠CBD
∵BA=BC、BD=BD
∴△ABD≌△CBD (SAS)
∴∠ADB=∠CDB
∵∠ADP=180-∠ADB、∠CDP=180-∠CDB
∴∠ADP=∠CDP
∵PM⊥AD,PN⊥CD
∴∠PMD=∠PND=90
∵DP=DP
∴△PDM≌△PDN (AAS)
∴PM=PN
证明:∵BD是∠ABC的平分线,BA=BC, BD=BD
∴△ABD≌△CBD
∴∠ADB=∠CDB
∴∠ADP=∠CDP
又因为DP=DP PM⊥AD,PN⊥CD,
∴△PMD≌△PND
所以PM=PN.