解：?$(1)①∠CPB=180°-90°-∠EPD=90°-∠EPD=∠DEP$?
又?$∠D=∠C=90°$?
∴?$△BPC∽△PED$?
?$②∠BPC=90°-∠CPE=∠PEC，$??$∠BCP=∠PCE=90°$?
∴?$△BPC∽△PEC$?或?$△BEP∽△BPC$?
?$③∠BPC=90°-∠PBC=∠EBP，$??$∠C=∠EPB=90°$?
∴?$△BPC∽△EBP$?
?$(2)①\frac {PD}{BC}=\frac 12$?
∴?$△PED$?與?$△BPC$?的周長比是?$\frac 12$?
?$②\frac {PC}{DC}=\frac 12$?
∴?$△PEC$?與?$△BPC$?的周長比是?$\frac 12$?
?$△BEP$?與?$△BPC$?的周長比是?$\frac {\sqrt 5}2$?
?$③BP=\sqrt {BC^2+PC^2}=\sqrt 5PC，$??$\frac {BP}{PC}=\frac {\sqrt 5}1$?
∴?$△EBP$?與?$△BPC$?的周長的比是?$\frac {\sqrt 5}1$?
