23.如圖,AB=12cm,AC⊥AB,BD⊥AB,AC=BD=9cm,點(diǎn)P在線(xiàn)段AB上以3cm/s的速度,由A向B運(yùn)動(dòng),同時(shí)點(diǎn)Q在線(xiàn)段BD上由B向D運(yùn)動(dòng).
(1)若點(diǎn)Q的運(yùn)動(dòng)速度與點(diǎn)P的運(yùn)動(dòng)速度相等,當(dāng)運(yùn)動(dòng)時(shí)間t=1(s),△ACP與△BPQ是否全等?說(shuō)明理由,并直接判斷此時(shí)線(xiàn)段PC和線(xiàn)段PQ的位置關(guān)系;
(2)將“AC⊥AB,BD⊥AB”為改“∠CAB=∠DBA”,其他條件不變.若點(diǎn)Q的運(yùn)動(dòng)速度與點(diǎn)P的運(yùn)動(dòng)速度不相等,當(dāng)點(diǎn)Q的運(yùn)動(dòng)速度為多少時(shí),能使△ACP與△BPQ全等.
(3)在圖2的基礎(chǔ)上延長(zhǎng)AC,BD交于點(diǎn)E,使C,D分別是AE,BD中點(diǎn),若點(diǎn)Q以(2)中的運(yùn)動(dòng)速度從點(diǎn)B出發(fā),點(diǎn)P以原來(lái)速度從點(diǎn)A同時(shí)出發(fā),都逆時(shí)針沿△ABE三邊運(yùn)動(dòng),求出經(jīng)過(guò)多長(zhǎng)時(shí)間點(diǎn)P與點(diǎn)Q第一次相遇.