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(1)证明：因为直[文]线AB⊥平面BCD[章]E,所以AB⊥BC[来],AB⊥BE,因为[自]底面BCDE是梯形[知]，BC∥DE,BC[嘛]⊥CD,CD=DE[答]=2BC=2,F是[案]边BC的中点，所以[网]四边形CDEF为正[文]方形，BE=√EF[章]2十BF=2√2，[来]CE2=8,AE2[自]=AB2+BE2=[知]AB2+8,AC2[嘛]=AB2+BC2=[答]AB2+16,所以[案]AC2=AE2十C[网]E2,所以AE⊥E[文]C.(2)解：以F[章]E,FC所在直线分[来]别为x,y轴，在平[自]面ABC中，过点F[知]且垂直于FC的直线[嘛]为之轴，建立如图所[答]示的空间直角坐标系[案]Fxyz,设AB=[网]a,则C(0,2,[文]0),E(2,0,[章]0),D(2,2,[来]0),A(0,-2[自],a),FA=(0[知],-2,a),FD[嘛]=(2,2,0),[答]设平面ADF的一个[案]法向量为n=(x,[网]y,z),FA·n[文]=-2y十az=0[章],则所以n=(-a[来],a,2).FD·[自]n=2x+2y=0[知],E因为直线AB⊥[嘛]平面BCDE,所以[答]AB⊥CE,又EC[案]2十EB2=BC2[网],所以BE⊥EC,[文]因为BE∩AB=B[章],所以CE⊥平面A[来]BE,所以平面AB[自]E的一个法向量为C[知]E=(2,一2,0[嘛])，因为平面ADF[答]与平面ABE所成二[案]面角为45°，所以[网]cos(CE,n)[文]1=cE·nAa=[章]c0s45°=CE[来]1|n|22X√2[自]a2+4,解得a=[知]√2，所以AD=([嘛]2,4,一√2).[答]设直线AD与平面A[案]BE所成的角为0，[网]则sin0=cos[文](AD,C克1=1[章]A方.CE4ADI[来]CE22X2211[自]所以直线AD与平面[知]ABE所成角的正弦[嘛]值为√1I11
One possible version:D[答]ear Jack,The Dragon Boat Festival,[案]which falls on the 5th day of the 5th lunar month,is one of the most im-portan[网]t tradition[文]al festivals[章].Every year Chinese people will hold grand celebrati[来]ons in memory of QuYuan.Kn[自]owing that you are intereste[知]d in Chinese tradition[嘛]s,I'm writing to invite you to take part in thecelebr[答]ation to be held by our school.Th[案]e celebrati[网]on will take place by the lake of our school on Friday.In[文] the morning,w[章]e will wrapand cook "zongzi",which will be eaten as lunch.In the afternoon[来],a dragon boat race will be held on thelake.I[自]'m sure you will have a good time.Wait[知]ing for your coming.Yo[嘛]urs,Li Hua
