a) Let $D$ be the origin and let the side length of the square be $2a$. then $E=(a,2a)$, $F=(0,a)$, and $C=(2a,0)$. Therefore the slope of $ED$ is $2$ and the slope of $FC$ is $-\frac{1}2$. Since the slopes are negative reciprocals of each other, $ DE\perp CF $ b)We keep $2a$ as the side length of the square. Using result a), we find that $\triangle{FDP}\sim\triangle{FCD}\sim{DCP}$. Thus $DP=\frac{(FP)(DC)}{DF}=\frac{2a(FP)}a=2FP$ and so $PC=\frac{DP^2}{(FP)}=\frac{(2FP)^2}{(FP)}=4FP$ and $CF=FP+PC=5FP$. We also have $\triangle{FDP}\sim\triangle{EDA}$ and the scale factor is $\frac{ED}{FD}=\frac{a\sqrt{5}}a=\sqrt{5}$. Thus $DE=\sqrt{5}(AE)=\sqrt{5}[\sqrt{5}(FP)]=5FP$, and it follows $EP=DE-DP=5FP-2FP=3FP$ Finally we conclude $CF:PC:EP=5FP:4FP:3FP=5:4:3$