如图,三角ABC中,角B=角C,AB=AC=12cm,BC=8cm,点D是线段AB的中点,点P和Q是线段BC和AC上的动点,点P以每秒

如图,三角ABC中,角B=角C,AB=AC=12cm,BC=8cm,点D是线段AB的中点,点P和Q是线段BC和AC上的动点,点P以每秒2cm的速度从点B出发,沿线段BC的方向运动;同时点Q从点C出发,沿线段CA的方向运动,已知AB=AC,角B=角C,求:若点Q的运动速度与点P的运动速度相等,当运动时间t=1秒时,试比较线段DP与PQ的大小关系并说明理由?

解:
(1)当t=1时,BP=2cm CP=6cm CQ=2cm,因为D是AB中点所以BD=AD=6cm,因为∠B=∠C,BP= CQ=2cm ,BD=CP=6cm。所以△DBP全等于△PCQ(SAS),所以DP=PQ
(2)设点Q速度为x,则t秒后CQ长度为x cm,因为P的速度为2cm每秒,所以t秒后BP长度为2t cm CP=8-2t(cm)。当DB=CP,∠B=∠C,BP=CQ时,△DBP全等于△PCQ(SAS),列方程解得t=1,x=2。当DP=CQ,∠B=∠C,BP=CP时,,△DBP全等于△QCP(SAS),列方程解得t=2秒,x=3cm。
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第1个回答  2011-06-10
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第2个回答  2011-06-22
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第3个回答  2011-06-25
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