證明：??$(1)∵$??四邊形??$ABCD$??是正方形，
??$∴∠DAB=∠B=∠D=∠DCB=90°，$????$AB=BC=CD=AD，$??
??$∴DA$??和??$CD$??都是圓??$B$??的切線.
??$∵PQ_{切圓}B$??于??$F，$??
??$∴AP=PF，$????$QF=CQ，$??
??$∴△DPQ$??的周長(zhǎng)是??$DP+DQ+PQ=DP+DQ+PF+QF$??
??$=DP+AP+DQ+CQ=AD+CD=8.$??
∵正方形??$ABCD$??的周長(zhǎng)??$=4×4=16，$??
??$∴△DPQ$??的周長(zhǎng)等于正方形??$ABCD$??的周長(zhǎng)的一半.
??$(2)$??解：∵在??$Rt△PDQ $??中，??$DP^2+DQ^2=PQ^2，$??
??$∴(4-x)^2+(4-CQ)^2=(x+CQ)^2，$??
解得??$CQ=\frac {16-4x}{x+4}，$??
??$∴DQ=4-\frac {16-4x}{x+4}=\frac {8x}{x+4}.$??
∵四邊形??$ABCD$??是正方形，
??$∴AD∥BC，$??
??$∴△PDQ∽△MCQ，$??
??$∴\frac {DP}{CM}=\frac {DQ}{CQ}，$??即??$\frac {4-x}{y-4}=\frac {\frac {8x}{x+4}}{\frac {16-4x}{x+4}}，$??
整理可得??$y=\frac {8}{x}+\frac {1}{2}x.$??
因此??$y$??與??$x$??之間的函數(shù)關(guān)系式是??$y=\frac {8}{x}+\frac {1}{2}x.$?
?
