(1)△APC∽△PBD,理由如下:
∵PC=PD=CD,∴△PCD是等边三角形,∴∠CPD=∠PCD=∠PDC=60°
∴∠APC+∠A=60°
又∠APB=120°,∴∠APC+∠DPB=60°
∴∠A=∠DPB,又∠PCD=∠PDC=60°
∴∠PCA=∠PDB=120°
∴△APC∽△PBD
(2)△APC∽△PBD∽△PAB
(1)△APC∽△PBD,理由如下:
∵PC=PD=CD,∴△PCD是等边三角形,∴∠CPD=∠PCD=∠PDC=60°
∴∠APC+∠A=60°
又∠APB=120°,∴∠APC+∠DPB=60°
∴∠A=∠DPB,又∠PCD=∠PDC=60°
∴∠PCA=∠PDB=120°
∴△APC∽△PBD
(2)△APC∽△PBD∽△PAB