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本期问题初339如图1,P、Q分别是正方形ABCD边AB、BC上的点,且AP=CQ,⊙O是过C、D、Q三点的圆,PE与⊙O切于点E.证明:PE=BQ.初340已知在一条直线l上的两点A、B之间的距离为10 000 cm,在A、B处各有一个可移动的挡板,并分别设为1号板、2号板.假设在A和B之间有一个乒乓球,且沿着直线l朝一个方向运动时是匀速的.开始时,1号板和乒乓球同时从点A处沿着直线l向点B的方向匀速运动,且速度分别为4 cm/s,5 cm/s,并假设1号板的速度始终保
The issue of the beginning of 339 Figure 1, P, Q are the ABCD AB, BC, the point on the side, and AP = CQ, ⊙ O over C, D, Q three o’clock, PE and ⊙ O cut at the point E. Proof: PE = BQ. The first 340 is known to have a distance of 10 000 cm between two points A and B on a straight line l and a movable baffle at points A and B, respectively Plate 1, Plate 2. Assume that there is a ping pong ball between A and B and is uniform in one direction along line I. Initially, plate 1 and ping pong ball are simultaneously run from point A Straight line l to the direction of point B uniform motion, and the speed were 4 cm / s, 5 cm / s, and assuming that the speed of plate 1 always