Because AB=AC, ∠ABC=∠ACP

Because BD=PQ, CQ=BP.

According to the corner theorem

So triangle BPD is congruent with triangle CQP.

2. If the triangle BPD is congruent with the triangle CQP, only the sides BD=CP, BP=CQ or BD=QC, BP=CP, because ∠ABC=∠ACP are equal, and the sides DP and PQ are variable.

So there are two cases, let the moving speed of point Q be Xcm/ s, and the two triangles are congruent after y seconds.

The first case:

When BD=CP and BP=CQ, the following equation can be listed:

5=8-3Y (listed according to BD=CP)

3Y=XY (listed according to BP=CQ)

If X=3 and Y= 1 are solved, the moving speed of point q is 3cm/ s, which is inconsistent with the equal speed of point p.

The second case:

When BD=QC and BP=CP, the following equation can be listed:

5=XY

3Y=8-3Y

Solve X= 15/4 and Y=4/3, that is, point q moves for 4/3 seconds at the speed of 15/4cm/ s per second, so that triangle BPD and triangle CQP can be congruent.

3. As can be seen from the meaning of the question, the speed of point Q is 15/4cm/ s, and the speed of point P is 3cm/ s, which means that point P is chased counterclockwise by Q, and the original counterclockwise distance between point C and point B is10+/kloc-0 = 20cm, so 20 ÷ (.

Then walk 80cm counterclockwise from point B, that is, turn 84cm three times, and then return to point B, that is, the side of AB.

So after 80/3 seconds, point P and point Q meet for the first time on the AB side of △ABC.