边长为a的正三角形ABC内一点P,分别向三条边作垂线得到PD,PE,PF,三角形的高为h,

(1)S=√3/4*a^2
(2)h=PD+PE+PF=√3/2*a
(3)h=1,a=2√3/3
PD=1/2,PE=1/3,PF=1-PD-PE=1-1/2-1/3=1/6
h/a=sin60°=√3/2
h=1,a=1/(√3/2)=2√3/3
PD=1/2,PE=1/3,PF=1-PD-PE=1-1/2-1/3=1/6
∵PE=1/3h
过P作PM//AC交AB于M
∴AM/AB=PE/h=1/3
∴AM=1/3AB=2√3/9
∴∠PMN=∠A=60°
又∵PD=1/2h
过P作PN//BC交AB于N
∴BN/BA=PD/h=1/2
∴BN=1/2AB=√3/9
∴∠PNM=∠B=60°
∴△PMN是等边三角形
∵PF⊥MN
∴F是MN中点
MF=NF=1/2MN=1/2(AB-1M-BN)
=√3/18
∴AF=AM+MF=5√3/18
BF=BN+NF=7√3/18
同理可得：AE=2√3/9,CE=4√3/9
BD=5√3/18,CD=7√3/18
S△AEF=1/2·AE·AF·sinA
=1/2×2√3/9×5√3/18·sin60°
=5√3/108
S△CDE=1/2·CD·CE·sinC
=1/2×7√3/18×4√3/9·sin60°
=7√3/54
S△BDF=1/2·BD·BF·sinB
=1/2×5√3/18×7√3/18·*sin60°
=35√3/432
S△ABC=1/2·a·h
=1/2×2√3/3×1
=√3/3
S△DEF=S△ABC-S△AEF-S△CDE-S△BDF
=√3/3-5√3/108-7√3/54-35√3/432
=11√3/432