郭敦顒回答：
∵在Rt⊿ABC中,∠BAC=90°,AD⊥BC于D,∠ABC的平分线分别交AD、AC于点E、F,点H是EF的中点.
（1）∵BF是∠ABC的平分线,∴∠ABF=∠DBE,
在Rt⊿BAF与Rt⊿BDE中,∠BAF=∠BDE中=90°,
∴∠AFB=∠DEB（两△中,两对应角相等,则第三角相等）,
∵∠AFB=∠AFE（同角）,∠DEB=∠AEF（对顶角）,
∴∠AFE=∠AEF,∴AF=AE,
∴△AEF是等腰三角形.
（2）在Rt⊿BAF与Rt⊿BDE中,
∵BF是∠ABC的平分线,∴∠ABF=∠DBE,
∴Rt⊿BAF∽Rt⊿BDE.
（3）设△AHF、△BDE、△BAF的周长分别为C1 C2 C3.
求证 （C1+C2）/C3≤9/8,并求出当等号成立时式子AF/BE的值.
验证：
当AB=AC时,并令AB=AC=1,则∠ABF=∠FAH=22.5°,
∴BD=（1/2）√2=0.70710678
AE=AF=1•tan22.5°=0.41421356,
BF=1/ cos22.5°=1.0823922
AH=AF•cos22.5°=0.41421356×0.9238795=0.3826834
FH= AF•sin 22.5°=0.41421356×0.38268343=0.15851267
DE=（1/2）√2－AE=0.70710678－0.41421356=0.2928932
BE=BF－2FH=1.0823922－2×0.15851267=0.76536686
C1= AH+ FH+AF=0.3826834+0.15851267+0.41421356=0.95540963
C2=BD+DE+BE=0.70710678+ 0.2928932+0.76536686=1.765366845
（C1+C2）=0.95540963+1.765366845=2.720776476
C3=AB+AF+BF=1+0.41421356+1.0823922=2.49660576
（C1+C2）/C3=2.720776476/2.49660576=1.08979＜ 9/8=1.125
当∠ABC=60°,∠ABF=∠CBF=∠ACB=∠FAH=30°时,令AB=1,则
∴BD=1/2=0.5
AE=AF=1•tan30°=0.57735,
BF=1/ cos30°=1.1547,
AH=AB•sin 30°=0.5,
FH= AF/2=0.288675,
DE=0.5×tan30°=0.288675,
BE=2DE=0.57735,
C1= AH+ FH+AF=0.5+0.288675+0.57735=1.366025,
C2=BD+DE+BE=0.5+ 0.288675+0.57735,=1.366025,
（C1+C2）=1.366025+1.366025=2.73205,
C3=AB+AF+BF=1+0.57735+1.1547=2.73205
（C1+C2）/C3=2.73205/2.73205=1 ＜ 9/8=1.125
当∠ABC=30°,∠ABF=∠CBF=∠FAH=15°时,令AB=1,则
∴BD=0.8660254,AD= 0.5,
AE=AF=1•tan15°=0.2679492,
BF=1/ cos15°=1.035276,
AH=1•sin 15°=0.258819,
FH= AF•sin 15°=0.2679492×0.258819=0.06935,
DE=0.5－AE=0.5－0.2679492==0.2320508,
BE=BF－2FH=1.035276－2×0.06935=0.896576,
C1= AH+ FH+AF=0.258819,+0.06935+0.2679492=0.5961182,
C2=BD+DE+BE=0.8660254+ 0.2320508+0.896576=1.9946522,
（C1+C2）=0.5961182+1.9946522=2.5907704,
C3=AB+AF+BF=1+0.2679492+1.035276=2.303225
（C1+C2）/C3=2.5907704/2.303225≈1.125 = 9/8=1.125
∵AF=1•tan15°=0.2679492,
BE=BF－2FH=1.035276－2×0.06935=0.896576,
∴AF/BE=0.2679492/0.896576=0.298858.