(1)∵将△ABC绕点B按逆时针方向旋转,得到△A1BC1,∴∠A1C1B=∠ACB=45°,BC=BC1,
∴∠CC1B=∠C1CB=45°,
∴∠CC1A1=∠CC1B+∠A1C1B=45°+45°=90°;
(2)∵△ABC≌△A1BC1,
∴BA=BA1,BC=BC1,∠ABC=∠A1BC1,
∴
| BA |
| BC |
| BA1 |
| BC1 |
∴∠ABA1=∠CBC1,
∴△ABA1∽△CBC1,
∴
| S△ABA1 |
| S△CBC1 |
| AB |
| BC |
| 4 |
| 5 |
| 16 |
| 25 |
∵S△ABA1=4,
∴S△CBC1=
| 25 |
| 4 |

