如图,圆柱的轴截面(过旋转轴的界面)ABCD是正方形,点E为弧AB的中点,AF垂直于DE,F是垂足

1.证：∵DA⊥平面AEB,∴DA⊥EB,∵E在半圆AB上,∴AE⊥EB,∴EB⊥平面ADE,
∵ EB在平面DEB内,∴平面ADE⊥平面DEB.
2.证：∵已证平面ADE⊥平面DEB,AF在平面ADE内,AF⊥DE,∴AF⊥平面DEB,∴AF⊥DB .
3.设正方形ABCD边长为1,作DB中点O,连结AO、FO、FB,
易得DA=1,DB=√2,AE=EB=√2/2,DE=√6/2,
∴AO⊥DB,Rt△DAE中,AF⊥DE,∴AF=AD*AE/DE=(1*√2/2)/(√6/2)= √3/3,
∵Rt△AFE中EF^2=AE^2-AF^2=(√2/2)^2-(√3/3)^2,∴EF=√6/6,
∵Rt△FEB中FB^2=FE^2+EB^2=(√6/6)^2+(√2/2)^2,∴FB=√6/3,
又DF=DE-FE=√6/2-√6/6=√6/3,∴DF=FB,∴FO⊥DB,Rt△AFO的锐角AOF为二面角A-DB-E
易知 △DOF∽ △DEB,∴ FO:BE=DF:DB,∴FO=BE*DF/DB=(√2/2)*(√6/3)/√2=√6/6,
∴ tanAOF=AF/FO=(√3/3)/(√6/6)=√2.