Question

矩形纸片ABCD中,AB=5,AD=3,将纸片折叠,使点B落在边CD上的B'处,折痕为AE,折痕为AE,

Answer 1

AD=3,AB′=4,DB′=√7

BE=EB′=x

CE=3-x,CB′=4-√7

x²=（3-x）²+（4-√7）²

x=（16-4√7）/3、

PB′‖BC

∠BEP=∠EPB′

∠BEP=∠B′EP

∠EPB′=∠B′EP

PB′=B′E=BE==（16-4√7）/3