圆O内两等弦AB、CD相交于点E,且AD、CB的延长线相交于点P.求证:1.AP=CP 2.PD=PB
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![圆O内两等弦AB、CD相交于点E,且AD、CB的延长线相交于点P.求证:1.AP=CP 2.PD=PB](/uploads/image/z/1603003-67-3.jpg?t=%E5%9C%86O%E5%86%85%E4%B8%A4%E7%AD%89%E5%BC%A6AB%E3%80%81CD%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9E%2C%E4%B8%94AD%E3%80%81CB%E7%9A%84%E5%BB%B6%E9%95%BF%E7%BA%BF%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9P.%E6%B1%82%E8%AF%81%EF%BC%9A1.AP%3DCP+2.PD%3DPB)
圆O内两等弦AB、CD相交于点E,且AD、CB的延长线相交于点P.求证:1.AP=CP 2.PD=PB
圆O内两等弦AB、CD相交于点E,且AD、CB的延长线相交于点P.求证:1.AP=CP 2.PD=PB
圆O内两等弦AB、CD相交于点E,且AD、CB的延长线相交于点P.求证:1.AP=CP 2.PD=PB
孤对应的角是相等的,利用全等三角形就可以了.
证明:①∵⌒BD=⌒DB
∴∠DAB=∠DCB
∵⌒AC=⌒CA
∴∠ADC=∠ABC
∵∠ADP=∠CBP
∴∠ADP-∠ADC=∠CBP-∠ABC
∴∠ABP=∠CDP
...
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证明:①∵⌒BD=⌒DB
∴∠DAB=∠DCB
∵⌒AC=⌒CA
∴∠ADC=∠ABC
∵∠ADP=∠CBP
∴∠ADP-∠ADC=∠CBP-∠ABC
∴∠ABP=∠CDP
∵AB=CD
∴△ABP≌△COP(ASA)
∴AP=CP
②由①得
△ABP≌△COP
∴PD=PB
收起
证明:①∵⌒BD=⌒DB
∴∠DAB=∠DCB
∵⌒AC=⌒CA
∴∠ADC=∠ABC
∵∠ADP=∠CBP
∴∠ADP-∠ADC=∠CBP-∠ABC
∴∠ABP=∠CDP
...
全部展开
证明:①∵⌒BD=⌒DB
∴∠DAB=∠DCB
∵⌒AC=⌒CA
∴∠ADC=∠ABC
∵∠ADP=∠CBP
∴∠ADP-∠ADC=∠CBP-∠ABC
∴∠ABP=∠CDP
∵AB=CD
∴△ABP≌△COP(ASA)
∴AP=CP
②由①得
△ABP≌△COP
∴PD=PB
收起