Question

在三角形ACB中,∠ACB＝90°,D为BC的中点,FG平行AC,CF过AD点E,求证：BF=2CG.

Answer 1

∵∠CAD = 90°,E是AD的中点

∴EC = ED = EA

∴∠ECB = ∠GDC

∵AC//GF

∴EG：EA = FE：EC

∴EG：EF = EC;EA

∵∠CEG = ∠AEF

∴△GEC∽△FEA

∴∠GCE = ∠FAE

∵∠CFB = ∠FAE +∠EAC+∠ACE

∠CGD=∠EAC+∠ACE+∠ECG

∴∠CFB = ∠CGD

∴△CFB∽△DGC

∴CB：CD = BF：GC

∵D是CB的中点

∴CB：CD = 2：1

∴BF：CG = 2：1

∴2CG = BF