1 |
4 |
1 |
3 |
2 |
3 |
3 |
4 |
1 |
3 |
2 |
3 |
1 |
3 |
5 |
2 |
1 |
2 |
1 |
3 |
1 |
2 |
k |
x |
3 |
2 |
1 |
x+1 |
x |
x2+2x+1 |
k |
x |
3 |
5 |
如图,在正方形ABCD中,CE⊥DF.求证:CE=DF.证明:设CE与DF交于点O,∵四边形ABCD是正方形,∴∠B=∠DCF=90°,BC=CD.∴∠BCE+∠DCE=90°,∵CE⊥DF,∴∠COD=90°.∴∠CDF+∠DCE=90°.∴∠CDF=∠BCE,∴△CBE≌△DFC.∴CE=DF. |
EG |
FH |
EG |
FH |
CE |
BF |
PQ |
OQ |
PQ |
OQ |