引理:sinθ=2Skl
证明:S=S△EFG+S△EHG,
=S△EOF+S△GOF+S△EOH+S△GOH,
=12 EO•OF•sinθ+12GO•OF•sin(180°-θ)
+12EO•OH•sin(180°-θ)+12 GO•OH•sinθ
=12EG•OF•sinθ+12EG•OH•sinθ
=12EG•FH•sinθ=12kl•sinθ
所以sinθ=2Skl
过E、F、G、H分别作AB、BC、CD、DA的垂线,得矩形PQRT.
设正方形ABCD的边长为a,PQ=b,QR=c,
则b=根号(k2 -a2),c=根号(l2 -a2),
由S△AEH=S△TEH,S△BEF=S△PEF,S△GFC=S△QFG,S△DGH=S△RGH
得SABCD+SPQRT=2S,
∴a2+bc=2S,即a2+根号(k2-a2)•根号(l2-a2 ) =2S,
∴(k2+l2-4S)a2=k2l2-4S2,
由引理知kl=2Ssinθ>2S,
所以k2+l2≥2kl>4S,
故SABCD=a2=(k2l2-4S2)(k2+l2-4S)