
解:(1)連接AQ
∵AB∥PQ AB=PQ
∴四邊形ABPQ是平行四邊形,
∴AQ∥BP AQ=BP
∴△ADQ∽△CDB,
∵BP=3
∴AQ=3
∵

,
∴

,
∴

;
(2)∵AB∥PQ,AQ∥BC,
∴

,

∵AB=4,BC=5,AC=6,BP=x,DE=y,
當點P在邊BC上時,
∴

,解得

…

,解得

…
∴

…
當點P在邊BC的延長線上時,
∴

,解得

…

,解得

…
∴

綜上所述,

(x>0)…
(3)∵AB∥PQ,∴△EDQ∽△ADB …
又以Q、D、E為頂點的三角形與△ABC相似,
∴△ADB與△ABC相似 …
∵∠BAC公共,又∠ABD≠∠ABC
∴∠ABD=∠ACB …
∴

即

由(2)知,

∴

得x=4
所以,當BP為4時,以Q、D、E為頂點的三角形與△ABC相似.…
分析:(1)連接AQ,由平行四邊形的判定定理可得出四邊形ABPQ是平行四邊形,進而可得出△ADQ∽△CDB,由相似三角形的對應(yīng)邊成比例即可得出結(jié)論;
(2)由平行線分線段成比例定理可知

,

,再根據(jù)點P在邊BC上或點P在邊BC的延長線上兩種情況討論即可;
(3)先由相似三角形的判定定理得出△EDQ∽△ADB,△ADB∽△ABC,由相似三角形的對應(yīng)邊成比例即可求出BP的長.
點評:本題考查的是相似三角形的判定與性質(zhì)、平行四邊形的判定定理及平行線分線段成比例定理,在解(2)時要注意分類討論,不要漏解.