右边的E为D,
(1)
AD=BC=AB=CD=5
L1平行于L2平行于L3平行于L4,ABCD为正方形,
AB平行于CD,所以BFDE为平行四边形,DE=BF,BE=FD
AE=AB-BE=CD-BE=CD-FD=CF,
三角形AED全等于三角形CFB,[SSS]
(2)
作EM⊥BF于M,设BE=X;
作AG⊥BF于G,交ED于N,EM=NG=1;
AN=h,AG=AN+NG=h+1,
BE:BA=EM:AG
X:5=1:(h+1)
h=(5-X)/X
RT△ANE∽RT△DAE,
AE:ED=AN:AD
(AB-EB):ED=h:5
(5-X):ED=(5-X)/X:5
ED=5X;
角A=90°,
ED²=AE²+AD²
25X²=(5-X)²+5²
25X²=25-10X+X²+25
12X²+5X-25=0
(4X-5)(3X+5)=0
X1=5/4,X2=-5/3(舍去)
h=(5-X)/X=(5-5/4)/(5/4)=3
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