己知AD,BE,CF分别为锐角三角形ABC的三边上的高, 作DP⊥AB,DQ⊥AC，P,Q为垂足.求证:(1),PQ=(DE+DF+EF)/2；(2) 如果∠A为钝角，则PQ=(DE+DF-EF)/2.

(1)延长DP,DQ与直线EF交于M,
BE⊥AC,CF⊥AB==>A.E.H.F共圆==>∠AEF=∠AHF
∠AEF=∠NEC,∠AHF=∠DHC
==>∠NEC=∠DHC
BE⊥AC,AD⊥BC==>H.D.C.E共圆==>∠DHC=∠DEC
==>∠NEC=∠DEC
DQ⊥AC
==>DE=EN,DQ=NQ
同理:DF=MF,DP=MP
==>PQ=MN/2=(MF+EF+EN)/2=(DE+DF+EF)/2

(2)设BE,CF,DA延长线交于点H,延长DP,DQ与直线EF交于M
证明方法与(1)类似,略
区别:MN=FM+EN-EF