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八年级数学解答题一般
题目
如图①,如图,在四边形ABCDABCD中,AB=CDAB=CD,EEFF分别是BCBCADAD的中点,连接EFEF并延长,分别与BABACDCD的延长线交于点MMNN.

(1)(1)求证:BME=CNE\angle BME=\angle CNE
(2)(2)如图②,在四边形ADBCADBC中,ABABCDCD相交于点OO,AB=CDAB=CD,EEFF分别是BCBCADAD中点,连接EFEF,分别交DCDCABAB于点MMNN,判断OMN\triangle OMN的形状.
知识点:三角形、三角形的中位线定理、角平分线的性质、等腰三角形的判定定理、解直角三角形章节:第13章 三角形 / 13.1 三角形的概念

答案与解析

答案

(1)(1)证明:如图所示,连接BDBD,取BDBD的中点HH,连接HEHEHFHF

E\because EFF分别是ADADBCBC的中点,
HF\therefore HFHEHE分别是BCD\triangle BCDABD\triangle ABD的中位线,
HF\therefore HFCN,HECN,HEBMBMHF=12CDHE=12ABHF=\frac{1}{2}CD,HE=\frac{1}{2}AB
AB=CD\because AB=CD
HF=HE\therefore HF=HE
HEF=HFE\therefore \angle HEF=\angle HFE
HF\because HFCN,HECN,HEBMBM
HEF=BME\therefore \angle HEF=\angle BMEHFE=CNE\angle HFE=\angle CNE
BME=CNE\therefore \angle BME=\angle CNE
(2)(2)OMN\triangle OMN是等腰三角形;
证明:如图,取BDBD的中点HH,连接HEHEHFHF

E\because EFF分别是BCBCADAD的中点,
HF\therefore HFHEHE分别是ABD\triangle ABDBCD\triangle BCD的中位线,
HF\therefore HFAB,HEAB,HECDCDHF=12ABHE=12CDHF=\frac{1}{2}AB,HE=\frac{1}{2}CD
AB=CD\because AB=CD
HF=HE\therefore HF=HE
HFE=HEF\therefore \angle HFE=\angle HEF
HF\because HFAB,HEAB,HECDCD
HFE=ONM\therefore \angle HFE=\angle ONMHEF=OMN\angle HEF=\angle OMN
ONM=OMN\therefore \angle ONM=\angle OMN
OM=ON\therefore OM=ON
OMN\therefore \triangle OMN是等腰三角形.

解析

(1)(1)证明:如图所示,连接BDBD,取BDBD的中点HH,连接HEHEHFHF

E\because EFF分别是ADADBCBC的中点,
HF\therefore HFHEHE分别是BCD\triangle BCDABD\triangle ABD的中位线,
HF\therefore HFCN,HECN,HEBMBMHF=12CDHE=12ABHF=\frac{1}{2}CD,HE=\frac{1}{2}AB
AB=CD\because AB=CD
HF=HE\therefore HF=HE
HEF=HFE\therefore \angle HEF=\angle HFE
HF\because HFCN,HECN,HEBMBM
HEF=BME\therefore \angle HEF=\angle BMEHFE=CNE\angle HFE=\angle CNE
BME=CNE\therefore \angle BME=\angle CNE
(2)(2)OMN\triangle OMN是等腰三角形;
证明:如图,取BDBD的中点HH,连接HEHEHFHF

E\because EFF分别是BCBCADAD的中点,
HF\therefore HFHEHE分别是ABD\triangle ABDBCD\triangle BCD的中位线,
HF\therefore HFAB,HEAB,HECDCDHF=12ABHE=12CDHF=\frac{1}{2}AB,HE=\frac{1}{2}CD
AB=CD\because AB=CD
HF=HE\therefore HF=HE
HFE=HEF\therefore \angle HFE=\angle HEF
HF\because HFAB,HEAB,HECDCD
HFE=ONM\therefore \angle HFE=\angle ONMHEF=OMN\angle HEF=\angle OMN
ONM=OMN\therefore \angle ONM=\angle OMN
OM=ON\therefore OM=ON
OMN\therefore \triangle OMN是等腰三角形.

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