Proste AB EF i CD są równoległe
Wykaż że
| 1 | 1 | ! | |||
= | + | ||||
| EF | AB | CD |
| EF | EC | ||
= | |||
| AB | AC |
| EF | AE | ||
= | (+) | ||
| CD | AC |
| EF | EF | EC | AE | ||||
+ | = | + | |||||
| AB | CD | AC | AC |
| EF*CD+EF*AB | EC+AE | ||
= | |||
| AB*CD | AC |
| EF(CD+AB | EC+AE | ||
= | |||
| AB*CD | AC |
| EF(CD+AB) | ||
więc | =1 | |
| AB*CD |
| EF | EF | EC + AE | |||
+ | = | ||||
| AB | CD | AC |
| 1 | 1 | |||
EF * ( | + | ) = 1 // :EF | ||
| AB | CD |
| 1 | 1 | 1 | |||
+ | = | ![]() | |||
| AB | CD | EF |