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電子課本網(wǎng) 第60頁

第60頁

信息發(fā)布者:
?$證明:連接?EG、??EH、??FG、??FH,?$?
?$∵?E?為?AD?的中點(diǎn),?H?為?AC?的中點(diǎn),$?
?$∴?EH∥CD,??EH=\frac {1}{2}CD,?$?
?$同理,?GF∥CD,??GF=\frac {1}{2}CD,?$?
?$∴?EH∥GF,??EH=GF,?$?
?$∴四邊形?EGFH?為平行四邊形,$?
?$∴?EF?與?GH?互相平分.$?

?$ 解:BC=2CF,理由如下:$?
?$∵D,E是AB,AC的中點(diǎn)$?
?$∴DE=\frac{1}{2}BC,DE//BF$?
?$∵EF//DC$?
?$∴四邊形CDEF是平行四邊形$?
?$∴DE=CF$?
?$∴BC=2CF$?
?$證明:?(1)?∵四邊形?ABCD?是平行四邊形,$?
?$∴?AB//CD,?$?
?$∴?∠GAE=∠HCF,?$?
?$∵點(diǎn)?G,??H?分別是?AB,??CD?的中點(diǎn),$?
?$∴?AG=CH,?$?
?$∵?AE=CF,?$?
?$∴?△AGE≌△CHF(\mathrm {SAS}),?$?
?$∴?GE=HF,??∠AEG=∠CFH,?$?
?$∴?∠GEF=∠HFE,?$?
?$∴?GE//HF,?$?
?$又∵?GE=HF,?$?
?$∴四邊形?EGFH?是平行四邊形.$?
?$?(2)?連接?BD?交?AC?于點(diǎn)?O,?$?
?$∵四邊形?ABCD?是平行四邊形,$?
?$∴?OA=OC,??OB=OD,?$?
?$∵?BD=14,?$?
?$∴?OB=OD=7,?$?
?$∵?AE=CF,??OA=OC,?$?
?$∴?OE=OF,?$?
?$∵?AE+CF=EF,?$?
?$∴?2AE=EF=2OE,?$?
?$∴?AE=OE,?$?
?$又∵點(diǎn)?G?是?AB?的中點(diǎn),$?
?$∴?EG?是?△ABO?的中位線,$?
?$∴?EG=\frac {1}{2}OB=3.5.?$?
?$∴?EG ?的長為?3.5?$?

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