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
答
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
)2=AG AD
,AG2
AD2
而
=S△ABG S△DBA
,∴BG BD
=AG2
AD2
.BG BD