1.已知等腰三角形ABC中,AB=AC=30cm,∠ABC=30°,CD⊥AB交BA延長(zhǎng)線于點(diǎn)D,AF為CA的延長(zhǎng)線,點(diǎn)P從A點(diǎn)出發(fā)以每秒1cm的速度在射線AF上向右運(yùn)動(dòng),連接BP,以BP為邊,在BP的左側(cè)作等邊三角形BPE,連接AE.
(1)如圖1,當(dāng)BP⊥AC時(shí),求證:△ABP≌△ACD;
(2)當(dāng)點(diǎn)P運(yùn)動(dòng)到如圖2位置時(shí),此時(shí)點(diǎn)D與點(diǎn)E在直線AP同側(cè),求證:AP=AB+AE;
(3)在點(diǎn)P運(yùn)動(dòng)過程中,連接DE,當(dāng)點(diǎn)P運(yùn)動(dòng)
秒時(shí)(直接寫答案),線段DE長(zhǎng)度取到最小值.
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