(1) Make bc parallel lines after crossing P, and cross ab to M.
Across cd to n, because the angle npb is the external angle of △bmp, ∴∠npq+∠qpb=∠mbp+∠pmb, ∫∠PMB =∠qpb = 90∞.
(2) The area of the quadrilateral PBCQ is equal to pqc+pbc= half *cq*pn+ half *bc*mb.
Because ap=x, cp= (root number two -x)
Pn= radical number two -x), and the rest are cq.
B.C.
Bm can be expressed by substituting s= half *cq*pn+ half *bc*mb to get the relation.
(3) This should be discussed in different categories. You can use X to represent both sides of an isosceles triangle.