As shown in the figure, P is the point on the fixed length line AB, and C and D respectively start from P and B and move to the left along the straight line AB at the speed of 1cm / s and 2cm / S (C is on the line AP) As shown in the figure, P is the point on the fixed length line AB, and C and D respectively start from P and B and move to the left along the line AB at the speed of 1cm / s and 2cm / S (C is on the line AP and D is on the line BP) (3) Under the condition of (1), if CD = 1 / 2Ab happens to exist after C and d move for 5 seconds, then point C stops moving and point d continues to move (point D is on line Pb), and m and N are the midpoint of CD and PD respectively. This paper attempts to explore the quantitative relationship between line Mn and AB, and gives a conclusion and reasons

As shown in the figure, P is the point on the fixed length line AB, and C and D respectively start from P and B and move to the left along the straight line AB at the speed of 1cm / s and 2cm / S (C is on the line AP) As shown in the figure, P is the point on the fixed length line AB, and C and D respectively start from P and B and move to the left along the line AB at the speed of 1cm / s and 2cm / S (C is on the line AP and D is on the line BP) (3) Under the condition of (1), if CD = 1 / 2Ab happens to exist after C and d move for 5 seconds, then point C stops moving and point d continues to move (point D is on line Pb), and m and N are the midpoint of CD and PD respectively. This paper attempts to explore the quantitative relationship between line Mn and AB, and gives a conclusion and reasons

(1) 2 (AP - t) = Pb – 2t2ap = BP, P point is one-third of a on the line AB (2) when q is ab trisection point, PQ / AB = 1 / 3, when q is on the extension line of AB, PQ = AB, PQ / AB = 1 (3) let AB length be a, AC = 1 / 3-5, DB = 10,1 / 3a-5 + 10 = 1 / 2a, a = 30, then AC = 5, Mn = 1 / 2 (cp-pd) = 2.5