如图,梯形ABCD中,AD∥BC,EF经过梯形对角线的交点O,且EF∥AD. (1)求证:OE=OF, (2)求OE/AD+OE/BC的值; (3)求证:1/AD+1/BC=2/EF.

问题描述:

如图,梯形ABCD中,AD∥BC,EF经过梯形对角线的交点O,且EF∥AD.

(1)求证:OE=OF,
(2)求

OE
AD
+
OE
BC
的值;
(3)求证:
1
AD
+
1
BC
2
EF

(1)∵EF∥AD,AD∥BC,

OE
BC
=
AO
AC
=
OD
BD
=
OF
BC

故OE=OF;
(2)∵EF∥AD,AD∥BC,
OE
AD
=
BE
AB
OE
BC
=
AE
AB

OE
AD
+
OE
BC
=
AE+BE
AB
=
AB
AB
=1;
(3)由(2)得:OE(
1
AD
+
1
BC
)=1,又OE=OF=
1
2
EF,
2OE
EF
=1,
∴OE(
1
AD
+
1
BC
)=
2OE
EF

1
AD
+
1
BC
2
EF