(1) If point P moves from point B to point C at a speed of 3cm/ s on BC line, at the same time, point Q moves from point C to point A on CA line.

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; I don't have to do this question because I have worked it out, but it may be helpful to look at this question.

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?

(2) If point Q starts from point C at the speed of ② and point P starts from point B at the same time at the original speed, and both move counterclockwise along the triangle of △ABC, how long will it take for point P and point Q to meet at which side of △ABC for the first time?

Answer:

Proof: 1. ①△BPD?△CQP。

Because AB=BD AB= 10.

So ab = BD = 5 ∠ B = ∠ C.

Point p and point q move at the same speed and time.

So BP=3 CQ=3.

Because BC=8

So PC=5

At △BPD and △CQP,

BP=CP

∠B=∠C

BD=CQ

So △ bpd △ cqp.

② Because the moving speed of P Q point is the same when △ BPD △ CQP.

However, under given conditions, the moving speed of P Q point is different.

So you can only get △ BPD △ CPQ.

At this time BD=CQ=5 BP=CP=4.

Because the moving speed of point p is 3 cm/s.

Therefore, the movement time of P Q point is t=4/3s.

Therefore, the moving speed of point q is v = 5/4/3 = 5 * 3/4 =15/4 cm/s.

2. because VP < VQ

So it took Q 20cm more than P to catch up with P.

Therefore, let the exercise time be t.

3t+20= 15/4t。

The solution is t=80/3s.

The passing distance of point P is 80/3s*3cm/s=80cm.

Because 80cm=2*28+24

So point P moved 80cm from point B, and then moved 24cm after two laps.

Of course, we met next to AB.