1. The side length of square ABCD is 16 √ 2, the diagonal lines AC and BD intersect at point O, through o as od1 ⊥ AB at D1, through D1 as D1D2 ⊥ BD at point D2, through D2 as d2d3 ⊥ AB at D3, and so on, where od1 + d2d3 + d4d5 + d6d7 = - cm 2. Place a triangle ruler on the rectangle ABCD with length √ 3 and width 1, and slide on the diagonal at its right angle vertex P. one side of the right angle always passes through point B, and the other side intersects with the extension line of DC at Q, (1) What is the size relationship between the line segment PQ and the line segment Pb when the point q is on the edge DC (2) When the point q is on the extension line of side DC, is the conclusion of (1) still valid (3) When point P slides on line AC, △ PBC becomes an isosceles triangle? If possible, point out all the positions of Q where △ PBC becomes an isosceles triangle. If not, explain why

1. The side length of square ABCD is 16 √ 2, the diagonal lines AC and BD intersect at point O, through o as od1 ⊥ AB at D1, through D1 as D1D2 ⊥ BD at point D2, through D2 as d2d3 ⊥ AB at D3, and so on, where od1 + d2d3 + d4d5 + d6d7 = - cm 2. Place a triangle ruler on the rectangle ABCD with length √ 3 and width 1, and slide on the diagonal at its right angle vertex P. one side of the right angle always passes through point B, and the other side intersects with the extension line of DC at Q, (1) What is the size relationship between the line segment PQ and the line segment Pb when the point q is on the edge DC (2) When the point q is on the extension line of side DC, is the conclusion of (1) still valid (3) When point P slides on line AC, △ PBC becomes an isosceles triangle? If possible, point out all the positions of Q where △ PBC becomes an isosceles triangle. If not, explain why

（1）15√2
Step: according to the meaning of the question
AD=CD=BC=AB
Because AC and BD are diagonals of a square
So the diagonals are divided equally
OA=OB=OC=OD
According to Pythagorean theorem
AC^2=CD^2+AD^2=(16√2)^2+(16√2)^2=1024
So AC = BD = 32
OA=OB=OC=OD=32/2=16
Because od1 ⊥ ab
So aod1 = bod1 = 8 √ 2
So od1 ^ 2 = 16 ^ 2 - (8 √ 2) ^ 2 = 128
OD1=8√2
According to the same method
Each side of the back is half of the previous one
OD1+D2D3+D4D5+D6D7=15√2cm
2.(1)
Because bpq is a right angle,
So ∠ EPB + ∠ FpQ = 90 degrees
Because ∠ EPB + EBP = 90 degrees
Therefore, EPB = FpQ and PEB = PFQ = 90 degrees,
Therefore, △ PEB is similar to △ QFP
So PQ / Pb = PF / EB
Because EB = FC, PQ / Pb = PF / FC
Because △ CFP is similar to △ CDA, PF / FC = ad / DC = root 3 / 1
So PQ / Pb = root 3 / 1
Let's call it a day!