Question

正方形ABCD,AD=12,E是CD边上的动点,AE的垂直平分线FP交AD.AE.BC.于点F.H.G,交AB的延长线于

Answer 1

AE=√(144+m^2),AH=HE=0.5√(144+m^2)

Rt△AHF∽Rt△ADE

HF:DE=AH:AD

FH=(m/24)*√(144+m^2)

AF^2=(144+m^2)/4+m^2(144+m^2)/24^2

=(144+m^2)(144+m^2)/24^2

AF=(144+m^2)/24=6+m^2/24

DF=6-m^2/24

作GG1⊥AG于G,GF⊥AE

∴ ∠G1GF=∠DAE

又,AD=GG1

∴ Rt△GG1F≌Rt△DAE

∴GF=AE

HG=√(144+m^2)-(m/24)√(144+m^2)

=√(144+m^2)(24-m)/24

FH:HG=m/(24-m)

(2)

FH:HG=1/2

2m=24-m

m=6,

FD=6-m^2/24=6-36/24=6-3/2=9/2

BG=12-m-FD=12-6-9/2=3/2

AF=12-FD=15/2

∵ Rt△PBG∽Rt△PAF

BG:AF=PB:(AB+BP)

1/5=PB/(12+PB)

12+PB=5PB

4PB=12

PB=3