On a straight line, consider the points A. B, C and D in consecutive form so that they form a harmonic quatern; over AB, BC and AD, their midpoints are marked P, Q and R respectively. Calculate $\frac{AP}{BQ}-\frac{AQ}{CR}$ (Answer:$\frac{3}{2}$)
My progress: Harmonic Quatern implies $\frac{AB}{BC}=\frac{AD}{CD}$ and $\frac{2}{AC}=\frac{1}{AB}+\frac{1}{AD}$
Let $AB=a, BC=b, CD = c$
$AP =BQ = \frac{a}{2}, PQ=QR = \frac{b}{2}$
$AR=RD= \frac{a+b+c}{2}$
$\frac{AP}{PQ} = \frac{\frac{a}{2}}{\frac{b}{2}}=\frac{a}{b} \implies \frac{AP}{BQ}-\frac{AQ}{CR}=\frac{AB}{BC}-\frac{AQ}{CR}=\frac{a}{b}-\frac{AQ}{QR}(I)$
$AQ = AB+BQ = a+\frac{b}{2}=\frac{2a+b}{2}\\ CR = AR-AC = \frac{a+b+c}{2}-(a+b)=\frac{-a-b+c}{2}$
From(I): $\frac{a}{b}-(\frac{2a+b}{-a-b+c})=\frac{-a^2-ab+ac-2ab-b^2}{b(-a-b+c)}=\frac{-a^2-3ab+ac-b^2}{b(-a-b+c)}$
...I couldn't resolve...