Question

正方形ABCD中,EF分别为AB延长线上两点,BE=AB,BF=BD,DF交CE于G,求证CG=BC

Answer 1

∵ABCD是正方形,∴AB＝BC＝DC＝AD、∠CBE＝∠BAD＝90°.

∵BE＝AB、AB＝BC,∴BE＝BC,又∠CBE＝90°,∴CE＝√2AB＝√2DC.

∵AB＝AD、∠BAD＝90°,∴BD＝√2AB＝√2DC,又BF＝BD,∴BF＝√2DC,

∴EF＝BF－BE＝√2DC－AB＝√2DC－DC.

∵DC∥EF,∴△CDG∽△FEG,∴DC/EF＝CG/EG,∴DC/（√2DC－DC）＝CG/（CE－CG）,

∴CG/（√2DC－CG）＝1/（√2－1）,∴CG/（√2DC）＝1/√2,∴CG＝DC＝BC.