scsnake 寫到:相似形?
O是BC和AP的交點。
ACO=BPO (angle in same seg.)
CAO=PBO (angle in same seg.)
AOC=BOP (vert. opp. angle)
ACO similar to BPO (AAA)
AO/BO=AC/BP (corr. sides, similar triangle)
BP = AC*BO/AO
BAO=PCO (angle in same seg.)
ABO=CPO (angle in same seg.)
BOA=POC (vert. opp. angle)
BAO similar to PCO (AAA)
AO/CO=AB/PC (corr. sides, similar triangle)
PC = AB*CO/AO
PAC=CAO (common angle)
APC=CBA=60(angle in same seg.)
=ACO (base angle of iso. triangle)
ACP=180-PAC-APC(angle sum of triangle)
=180-CAO-ACO
=AOC
PAC similar to CAO (AAA)
PA/CA=AC/AO
PA=AC^2/AO
PB+PC
=AC*BO/AO + AB*CO/AO
=(AB/AO)(BO+CO)
=(AB)^2/AO
=AC^2/AO
=PA