( 1)

Because a two-point formula of (0,4) c (4,0) can establish a linear equation.

Therefore, the analytical formula of the straight line where AC is located is Y=-X+4.

(2)

Extend BA to point e so that AE=AB. At this time, it is easy to know E (4,8) and PB=PE to prove the congruence of triangles.

So PB+PO=PE+PO

From the fact that the sum of any two sides of a triangle is greater than the third side, it can be seen that when P is the intersection of EO and AC, PB+PO is the smallest.

Because e (4,8) o (0,0), we know that the analytical formula of EO is Y=2X.

At the same time, EO and AC can get X=4/3.

That is, when PB+PO is minimum, the coordinates of point P are (4/3, 8/3).

(3)

PB=PC+CQ

Prove:

ABC is an isosceles right triangle.

PB & gtAB = AC & gtPC, so you must find a point D on PB, so that PC=PD.

Because PQ is the bisector of the angle BPC, all triangles PCQ are equal to triangles PDQ(SAS is an edge axiom).

So CQ=QD, angle PDQ= angle PCQ=45 degrees.

So BDQ angle = 135 degrees.

Because PB is the bisector of the angle ABC, the angle DBQ=22.5 degrees.

Because the sum of the internal angles of the triangle is 180 degrees, the DQB angle =22.5 degrees.

So DBQ is an isosceles triangle, that is, DB=DQ.

So PB=PD+BD=PD+DQ=PD+CQ=PC+CQ.

Certificate of completion

thank you