(1)当PQ⊥AD时△ADQ与△APD都是等腰直角三角形
∴AD/AQ+AD/AP=cos45°+cos45°=√2
∴ 1/AQ+1/AP=√2/AD
(2)如图②,当PQ不与AD垂直时,(1)的结论仍成立
过P,Q分别作AD和AD的延长线的垂线 垂足分别为F,G
∴AE=APcos45°,AG=AQcos45°
∴AD-DF=(√2/2)AP,AD+DG=(√2/2)AQ
∴AD/AP-DF/AP=√2/2,AD/AQ+DG/AQ=√2/2
∴AD/AP+AD/AQ+DG/AQ-DF/AP=√2
∵DG/AQ=DG/(√2QG)=(√2/2)*DG/QG=√2/2(1/tan∠QDG)
DF/AP=DF/(√2PF)=(√2/2)*DF/PF=√2/2(1/tan∠PDF)
∴AD/AP+AD/AQ=√2
∴1/AQ+1/AP=√2/AD
(3)AF=APcos30°,AG=AQcos30°
∴AD-DF=(√3/2)AP,AD+DG=(√3/2)AQ
∴AD/AP-DF/AP=√3/2,AD/AQ+DG/AQ=√3/2
∴AD/AP+AD/AQ+DG/AQ-DF/AP=√3
∵DG/AQ=DG/(2QG)=(1/2)*DG/QG=1/2(1/tan∠QDG)
DF/AP=DF/(2PF)=(1/2)*DF/PF=1/2(1/tan∠PDF)
∴AD/AP+AD/AQ=√3
∴1/AQ+1/AP=√3/AD
∴n=√3