Let $<\mathrm{QPR}=x$ Let length side small square be a and large square be b If PA'A is a straight line $\mathrm{AB}/\mathrm{PC}=A\text{'}B\text{'}/\mathrm{PB}\text{'}$ In triangle PD'C' $\mathrm{PC}\text{'}=D\text{'}C\text{'}\cot x=a\cot x$

$\mathrm{Therefore}\mathrm{PB}\text{'}=a+a\cot x$

Therefore $A\text{'}B\text{'}/\mathrm{PB}\text{'}=$ $\frac{a}{a+a\cot x}$

$=$ $\frac{1}{1+\cot }$ In triangle PDC $\mathrm{PC}=\mathrm{DC}\cot x=b\cot x$

$\mathrm{Therefore}\mathrm{PB}=b+b\cot x$

$\mathrm{Therefore}\mathrm{AB}/\mathrm{PB}=$ $\frac{b}{b+b\cot x}$

$=$ $\frac{1}{1+\cot x}$

$\mathrm{AB}/\mathrm{PC}=A\text{'}B\text{'}/\mathrm{PB}\text{'}$ Therefore PAA' is a straight line