Proof: Because AB=AC, ∠ B = ∠ C.

When exercise time is 1S, BP=CQ=3.

D is the midpoint of AB, so BD=AB/2=5.

CP=BC-BP=8-3-5

BD=CP

At △BPD and △CQP

BP=CQ，∠B=∠C，BD=CP

So △ BPD △ CQP (SAS)

② If the speeds of P and Q are not equal, then BP≠CQ.

Therefore, if two triangles are congruent, BP=CP, BD=CQ=5.

BP=CP=BC/2=4

Therefore, the movement time is 4/3 seconds.

Q Motion speed: 5÷4/3= 15/4.

(2) Since both points rotate counterclockwise, P leads Q20 cm(Q when starting (Q will catch up with P).

Let the distance traveled by P at the first meeting be X cm, and the distance traveled by Q be X+20 cm.

X/3=(X+20)/( 15/4)

Cross multiplication: 15X/4=3X+24

3X/4=60

X=80

So it takes 80/3 seconds to meet.

80-2( 10+ 10+8)=24

18 < 24 < 28, so we met at AB.