Question

已知△ABC是等边三角形点P是AC上一点(点P不与A,C重合)PE⊥BC于点E在CB的延长线上截取BD=PA连接PD设P

Answer 1

1、n=1时pa=pc=db

做af垂直于bc交bc于f

则ce=1/2cf=1/4bc

be=bc-ce=3/4bc=3/4ac

bd=pa=1/2ac

eb:bd=3:2

2、∠EPD=60°

又∠EPC=30°

则dp⊥ac

db=ac-pc=ac-1/2(ac+ap)=ap

db=1/3ac

n=1/2

eb=bc-ce=ac-1/2pc=2/3ac

eb:bd=2:1