四边形ABCD,∠BCD=1350,AB⊥BC,A

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应该是135°吧.作BH垂直于AE交AE于H,作CG垂直于BH交BH于G.易知CEHG是矩形.∴CE=GH,EH=CG∵∠BCD=135°∴∠GBC=∠BCG=45°∴BG=CG=√BC^2/2=3√2∴EH=3√2∵∠GBC=45,AB⊥BC∴∠ABH=∠HAB=45°∴AH=BH=AE-EH=10-3√2A...