∵△ABE≌△AB′E(折叠)∴AB=AB′=3,那么CB′=AC-AB′=5-3=2BE=B′E∠ABE=∠AB′E=∠ABC=90°∴∠EB′C=90°∵勾股定理:∠ABC=90°,AC=5,AB=3,那么BC=4∴CE=BC-BE=4-BE∴Rt△B′CE中:勾股定理:CE²=CB′²+B′E²那么(4-BE)²=2²+BE²16-8BE+BE²=4+BE²8BE=12BE=12/8=1.5