∴∠BAC=60。

AD = AE，

∴∠AED=60 =∠CEP，

∴∠EPC=30。

Triangle BDP is an isosceles triangle.

∫△AEP is similar to△△△ BDP,

∴∠EAP=∠EPA=∠DBP=∠DPB=30，

∴AE=EP= 1.

∴ in Rt△ECP, EC EC =12ep =12; +02;

Let BD = BC = X.

In Rt△ABC, from Pythagorean theorem, we get:

(x+ 1)2=x2+(2+ 1)2，

X=4, that is BC = 4.

Pass point c as cf ‖ DP.

△ ade is similar to△△△ AFC.

∴ AEAC=ADAF, that is, AF=AC, that is, DF=EC=2.

∴BF=DF=2.

∫△BFC is similar to△△△ BDP,

∴ BFBD=BCBP=24= 12, which means BC = CP = 4.

∴tan∠BPD= ECCP=24= 12。

(3) DQ⊥AC of Q point after D point.

Then △DQE is similar to △PCE. Let AQ=a, then QE =1-a.

∴ QEEC=DQCP and tan∠BPD= 13,

∴DQ=3( 1-a).

∵ At Rt△ADQ, according to Pythagorean theorem, AD2=AQ2+DQ2.

Namely: 12=a2+[3( 1-a)]2,

The solution is a= 1 (truncated) A = 45.

∫△ADQ is similar to△△△ ABC.

∴adab = dqbc = aqac = 45 1+x = 45+5x。

∴ AB=5+5x4, BC =3+3x4.

∴ The circumference of triangle ABC y = AB+BC+AC = 5+5x4+3+3x4+1+x = 3+3x,

That is, y=3+3x, where x > 0.