∴ BP = CQ = 3× 1 = 3cm,
∫AB = 10cm, and point D is the midpoint of AB.
∴BD=5 cm
∫PC = BC∫BP,BC=8 cm,
∴ PC = 8 3 = 5 cm,
∴PC=BD.
AB = AC,
∴∠B=∠C,
At △BPD and △CQP,
∴△BPD≌△CQP.(SAS)
②∵vP≠vQ,∴BP≠CQ,
And ∵△BPD?△CPQ, ∠B=∠C, BP=PC=4cm, CQ=BD=5cm,
Point p, point q moves in seconds,
∴ cm/sec;
(2) suppose that point p and point q meet for the first time after x seconds,
From the meaning of the question, x = 3x+2x 10,
Solve.
∴ Point P*** has moved by ×3=80 cm.
∴80=56+24=2×28+24,
Point p and point q meet on the side of AB,
A second later, point P and point Q meet at the edge of AB for the first time. Hope to adopt.