八年级数学竞赛题,会的老师同学帮忙做做,这道题很重要,打得好我一定给高分。亲们不要让我等太久哦

2024-12-25 11:30:55
推荐回答(2个)
回答1:

∠FCN=45º
作FH⊥MN于H∵∠AEF=∠ABE=90º∴∠BAE +∠AEB=90º,∠FEH+∠AEB=90º∴∠FEH=∠BAE 又∵AE=EF,∠EHF=∠EBA=90º∴△EFH≌△ABE ∴FH=BE,EH=AB=BC,∴CH=BE=FH∵∠FHC=90º,∴∠FCH=45º

回答2:

解:过点F作FG垂直MN于Q
所以角FQE=90度
因为ABCD是正方形
所以AB=BC
角ABE=90度
因为AEFG是正方形
所以AE=EF
角AEF=90度
因为角AEF+角ABB+角FEQ=180度
所以角AEB+角FEQ=90度
因为角ABE+角BAE+角AEB=180度
所以角BAE+角AEB=角AEB+角FEQ=90度
所以角BAE=角FEQ
角ABE=角FQE=90度
AE=EF(已证)
所以直角三角形ABE和直角三角形EQF全等(AAS)
所以BE=FQ
AB=EQ
因为AB=BC=BE+EC
EQ=EC+CQ
所以CQ=FQ
所以三角形FQE是等腰三角形
所以三角形FQC是等腰直角三角形
所以角FCN=45度

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