Inverse theorem from Pythagorean theorem: 3 2+4 2 = 5 2.

So Δ ABC is a right triangle, so Δ ABC =1/2ab * ac = 6.

Because the area of PBCQ is equal to the area of quadrilateral PBCQ,

Δ δAPQ area+quadrilateral area PBCQ = sδABC

So the area of Δ apq =1/2sΔ ABC = 3, so sΔ apq: sΔ ABC = 3: 6 =1:2.

Because PQ‖BC, so ∠APQ=∠B, ∠ AQP = ∠ C.

So δδAPQ is similar to δδABC.

So (pa: ab) = square of sδ apq: sδ ABC.

So PA=2* radical number 2

(2) Because δδAPQ is similar to δδABC.

So AP: AB = AQ: AC, which means AP: 4 = AQ: 3.

So AQ=3/4*AP

Because the perimeter of PBCQ is equal to the perimeter of quadrilateral PBCQ.

So AP+AQ+PQ=BP+CQ+PQ+BC.

Namely: AP+AQ=BP+CQ+5.

AP+3/4*AP=4-AP+3-AQ+5

7/4AP=4-AP+3-3/4AP+5

7/4AP= 12-7/4AP

3.5AP= 12

AP=24/7