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Math moving point problem in junior two.
Note: CD=4√2.

①a is AM⊥BC in M, D is DN⊥BC in N,

∵ad∨BC, ∴ quadrilateral ADNM is a rectangle,

∴MN=AD=5,

∠ c = 45, CD=4√2 in rt δ cdn,

∴CN=DN=CD÷√2=4,

∴BM= 12-5-4=3,

∴AB=√(BM^2+AM^2)=5。

① When BP=AB=5 cm, △ABP is an isosceles triangle with AB as the waist, corresponding to t = 5s;;

② When AP=AB=5 cm, △ABP is an isosceles triangle with AB as the waist.

BP =[√(AB2-H2)]* 2 =[√(52-42)]* 2 = 6,t = 6s;

③ When AP=BP and BQ= 1/2AB=5/2,

δabm∽δpbq,ab/bp=bm/bq,bp=5×5/2÷3=25/6,∴t=25/6

(2) When t= 1s, PB= 1 cm, PE=BC/2- 1=5=CD, PADE is a parallelogram;

When t= 1 1s, Pb =1/cm, PC= 1 cm, PE=BC/2-PC=5=CD, PDAE is a parallelogram;

⑶de^2=ce^2+cd^2-2ce*cd*cos45 =6^2+(4√2)^2-2*6*(4√2)*cos45=20,

DE=4√5 cm,DE≠AD;

Similarly, AE≠AD, because all four sides of the diamond are equal, P, A, D and E cannot form a diamond;