CindySun0527
幼苗
共回答了21个问题采纳率:95.2% 举报
(1)OA=OB=1,AB=√2;
设坐标P(m/2,1/m),则 AF=Yp/sin45°=√2/m,BE=AB-AE=√2-√2(OA-OM)=√2-√2(1 -m/2)=√2m/2;
∴ AF*BE=(√2/m)*(√2m/2)=1;
(2)E点坐标(m/2,1 -m/2),tan∠BOE=(m/2)/(1 -m/2)=m/(2-m);
F点坐标(1 -1/m,1/m),tan∠AOF=(1/m)/(1 -1/m)=1/(m-1);
tan∠AFO=-tan(∠AOF+45°)=[1 +1/(m-1)]/[1*1/(m-1) -1]=m/(2-m)=tan∠BOE;
△BOE和△AFO中,∠AOF=∠BFO,∠OBE=∠FAO=45°,所以两三角形相似;
(3)tan∠AOE=(1 -m/2)/(m/2)=(2-m)/m;
tan∠EOF=tan(∠AOF-∠AOE)=[1/(m-1) -(2-m)/m]/{1+[1/(m-1)]*[(2-m)/m]}=(2-2m+m²)/(2-2m+m²)=1;
∴ ∠EOF=45°;
1年前
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