As shown in the figure, it is known that in △ABC, AB=AC= 10cm, BC=8cm, and point D is the midpoint of AB. If point P moves from point B to point C at a speed of 3cm/s on BC line and point Q moves from point C to point A on CA line at the same time.

(1) If the moving speed of point Q is equal to that of point P, whether △BPD and △CQP are the same after 1 s, please explain the reason.

(2) If the moving speed of point Q is not equal to that of point P, when the moving speed of point Q is what, can △BPD and △CQP be congruent?

Answer:

1, proving that:

1 sec later, BP=CQ=3.

Ab = communication ∴∠ B = ∠ C.

∫CP = BC-BP = 8-3 = 5

BD=5

∴BD=CP

∴△BPD≌△CQP

2. Excluding this type of constant velocity, there is another possibility, that is, the midpoint between P and BC and the midpoint between Q and AC.

At this point:

CQ=PQ=DP=BD

BP=CP

△BPD?△CQP

If this condition holds, the movement of P is 4, the speed is 3 and the time is 4/3.

Q to the midpoint of AC, then when Q moves, the time is also 4/3 and the speed is 5÷4/3=3.75.