Question

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？

Answer 1

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。

Answer 2

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Answer 3

dad

Answer 4

22