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