请老师不吝赐教,

来源:学生作业帮助网 编辑:作业帮 时间:2024/04/28 20:49:45
请老师不吝赐教,
xRn@RVDBq%ۤ3&@!*JDy v,@} ,|DT骿ul YysΜ9R}nm0oZM,km 9q7T&SƲp*f-+z֬5{EQ[T״iz@hBZ0Cph0,-l*-dBBD!2)c\RAEMq#"X@NiC]Q; ܖJL @]Y0ag!!fX4t! .)HLR ӱU&f9י m&$(Υ-(6Uf\{ku7{v]/7_nswדsk#jm%֞9 ~ZRj'OVaA|LgwAe ` ugG=g@H-{0a ^^9m(j/ O'q^e?hKau%◇B==x;3M '8xԷ6

请老师不吝赐教,
请老师不吝赐教,

请老师不吝赐教,
⑴∵Rt△AB'C'是由Rt△ABC绕点A顺时针旋转得到
∴AB=AB' ,AC=AC' , ∠BAC=∠B'AC'
∵∠BAC+∠C'AB=∠C'AC , ∠B'C'A+∠C'AB=∠B'AB
∴∠C'AC=∠B'AB
∴∠ABB'+∠AB'B=∠ACC'+∠AC'C
∵AB=AB' ,AC=AC'
∴∠ABB'=∠AB'B ,∠ACC'=∠AC'C
∴∠ACC'=∠ABB'
∵∠CEA=∠BEF
∴△ACE∽△FBE
⑵∵由⑴得△ACE∽△FBE
∴∠BEF=∠CEA,∠FBE=∠ACE
∵AC=AC′
∴∠ACC′=(180°-∠CAC')/2=(180°-β)/2=90°-β/2
在Rt△ABC中,∠ACC′+∠BCE=90°,即:(90°-β/2)+∠BCE=90°
∴∠BCE=β/2
∵∠ABC=α
∴当β/2=α,即:β=2α时,可得:∠ABC=∠BCE
∴CE=BE
∴△ACE≌△FBE (ASA)