(1)∵AB∥x轴,∴a=4,AB=a+4-(-2)=4+6=10,∴BP=5,P坐标为(a-1,a)即(3,4)
(2)由题意知∠AOE=∠BOP=∠BOF,设其为x,∠AOD:∠POA=90-x:90+x-2x,∴两角之比为1:1.
(3)因为PF平分∠BPO(已知),∠AOD=∠AOP(已证),∠POD=∠BPO,所以∠AOP=∠FPO,AO∥PF∴AP=OF,A到B(x+2,y-4),P(已求)到F为(5,0)
OK
(1)∵AB∥x轴,∴a=4,AB=a+4-(-2)=4+6=10,∴BP=5,P坐标为(a-1,a)即(3,4)
(2)由题意知∠AOE=∠BOP=∠BOF,设其为x,∠AOD:∠POA=90-x:90+x-2x,∴两角之比为1:1.
(3)因为PF平分∠BPO(已知),∠AOD=∠AOP(已证),∠POD=∠BPO,所以∠AOP=∠FPO,AO∥PF∴AP=OF,A到B(x+2,y-4),P(已求)到F为(5,0)
OK