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
的值;OE BC
(3)求证:
+1 AD
=1 BC
. 2 EF
答
(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
=1;AB AB
(3)由(2)得:OE(
+1 AD
)=1,又OE=OF=1 BC
EF,1 2
∴
=1,2OE EF
∴OE(
+1 AD
)=1 BC
,2OE EF
∴
+1 AD
=1 BC
.2 EF