(2014?崇明县二模)已知:如图,在△ABC中,AB=AC,点D、E分别是边AC、AB的中点,DF⊥AC,DF与CE相交于

2025-01-01 00:29:02
推荐回答(1个)
回答1:

(1)∵AB=AC,AD=

1
2
AC,AE=
1
2
AB,
∴AD=AE,
在△BAD和△CAE中,
AB=AC
∠BAD=∠CAE
AD=AE

∴△BAD≌△CAE.
∴∠ABD=∠ACE,
∵DF⊥AC,AD=CD,
∴AF=CF,
∴∠GAD=∠ACE,
∴∠GAD=∠ABD.
∵∠GDA=∠ADB,
∴△GDA∽△ADB.
AD
DB
=
DG
AD

∴AD2=DG?BD.

(2)∵
AD
DB
=
DG
AD
,AD=CD,
CD
DB
=
DG
CD

∵∠CDG=∠BDC,
∴△DCG∽△DBC.
∴∠DBC=∠DCG.
∵AB=AC,
∴∠ABC=∠ACB.
∵∠ABD=∠ACE,
∴∠ECB=∠DBC=∠DCG,
∴∠ECB=∠DCG.