如图，连接AE、BE，由弦切角定理可知，∠PCA=∠PAE，
则△PAC∽△PEA，得
| AC |
| AE |
| PC |
| PA |
同理，
| BC |
| BE |
| PC |
| PB |
∵PA=PB，
∴
| AC |
| AE |
| BC |
| BE |
即
| AC |
| BC |
| AE |
| BE |
在⊙O中，由△ACD∽△EBD，△AED∽△CBD，
可得
| AC |
| BE |
| AD |
| ED |
| AE |
| BC |
| ED |
| BD |
从而
| AC |
| BC |
| AE |
| BE |
| AD |
| BD |
即
| AC2 |
| BC2 |
| AD |
| BD |
| AC2 |
| BC2 |
| AD |
| BD |
如图，连接AE、BE，| AC |
| AE |
| PC |
| PA |
| BC |
| BE |
| PC |
| PB |
| AC |
| AE |
| BC |
| BE |
| AC |
| BC |
| AE |
| BE |
| AC |
| BE |
| AD |
| ED |
| AE |
| BC |
| ED |
| BD |
| AC |
| BC |
| AE |
| BE |
| AD |
| BD |
| AC2 |
| BC2 |
| AD |
| BD |