已知,如图,在梯形ABCD中,AD∥BC,∠BCD=90°,对角线AC、BD相交于点E,且AC⊥BD. (1)求证:CD2=BC•AD; (2)点F是边BC上一点,联结AF,与BD相交于点G,如果∠BAF=∠DBF,求证:AG2AD2=BG

问题描述:

已知,如图,在梯形ABCD中,AD∥BC,∠BCD=90°,对角线AC、BD相交于点E,且AC⊥BD.

(1)求证:CD2=BC•AD;
(2)点F是边BC上一点,联结AF,与BD相交于点G,如果∠BAF=∠DBF,求证:

AG2
AD2
BG
BD

证明:(1)∵AD∥BC,∠BCD=90°,
∴∠ADC=∠BCD=90°,
又∵AC⊥BD,∴∠ACD+∠ACB=∠CBD+∠ACB=90°,
∴∠ACD=∠CBD,
∴△ACD∽△DBC,

AD
CD
=
CD
BC

即CD2=BC×AD;
(2)方法一:
∵AD∥BC,∴∠ADB=∠DBF,
∵∠BAF=∠DBF,∴∠ADB=∠BAF,
∵∠ABG=∠DBA,
∴△ABG∽△DBA,
AG
AD
=
AB
BD

AG2
AD2
=
AB2
BD2

又∵△ABG∽△DBA,
BG
AB
=
AB
BD

∴AB2=BG•BD,
AG2
AD2
=
AB2
BD2
=
BG•BD
BD2
=
BG
BD

方法二:∵AD∥BC,∴∠ADB=∠DBF,
∵∠BAF=∠DBF,∴∠ADB=∠BAF,
∵∠ABG=∠DBA,∴△ABG∽△DBA,
S△ABG
S△DBA
=(
AG
AD
2=
AG2
AD2

S△ABG
S△DBA
=
BG
BD
,∴
AG2
AD2
=
BG
BD