Here, $AB = 2017 \text{ units}$.
$O$ is mid-point on $AB$.
$D$ is such an interior point to $\Delta ABC$, so that $\angle DAP = \angle DBQ$.
Now, What is $(OP - OQ) $ ?
So far, I have proceeded to the point that $\Delta APD$ and $\Delta BQD$ are similar.
The value $OP-OQ=0$, regardless of $AB$'s length. Using median's length formula in $\triangle{APB}$ and $\triangle{AQB}$, is suffices to prove that $$\frac{AP^2 + BP^2}{2} - \frac{AB^2}{4} = \frac{AQ^2 + BQ^2}{2} - \frac{AB^2}{4}$$ $$\Longleftrightarrow AP^2 + BP^2 = AQ^2 + BQ^2$$
Using pythagorean theorem, we can write $AP^2 = AD^2 - PD^2$, and $ BQ^2 = BD^2- DQ^2$. Writing $BP^2$ and $AQ^2$ using law of cosines in $\triangle{BDP}$ and $\triangle{ADQ}$, it only remains to show $$BD\times DP\times\cos(\widehat{BDQ}) = AD\times DQ\times\cos(\widehat{ADQ})$$ Which is obvious from $\triangle{APD} \sim \triangle{BDQ}$.
