如图,在△ABC和△DEF中,已知AB=DE,BC=EF,根据“SAS”判定△ABC≌△DEF,还需要添加的条件是( )

| A | ∠A=∠D |
| B | ∠B=∠E |
| C | ∠C=∠F |
| D | 以上三个均可以 |
如图,AB=AD,AC=AE,∠BAC=∠DAE,下列结论错误的是( )

| A | ∠B=∠D |
| B | ∠C=∠E |
| C | BC=DE |
| D | BC=AE |
| A | AB=A′B′,AC=A′C′,∠C=∠C′ |
| B | AB=A′B′,∠A=∠A′,BC=B′C |
| C | AC=A′C′,∠A=∠A′,BC=B′C′ |
| D | AC=A′C′,∠C=∠C′,BC=B′C′ |
如图,AB∥CD,BC∥AD,AB=CD,BE=DF,图中全等三角形的对数是( )

| A | 1 |
| B | 2 |
| C | 3 |
| D | 4 |
| A | 1<AD<7 |
| B | 2.5<AD<5.5 |
| C | 2<AD<14 |
| D | 5<AD<11 |