For this, configuration, note that we have \[\angle PAO=\angle PCO\implies \angle PAO+\dfrac{\pi}{4}=\angle PCO+\dfrac{\pi}{4}\implies \angle PCD=\angle QAB.\] Similarly, we can show that $\angle QBO=\angle QDO\implies \angle PDC=\angle ABQ.$ Using these two equalities, it is evident that $PCD$ and $BAQ$ are similar triangles. So, \[\frac{AQ}{AB}=\frac{CD}{DP}\implies \frac 12 AQ\cdot DP=\frac 12 AB^2\implies [APQD]=\frac 12[ABCD];\] Where $[.]$ denotes the area, and $[APQD]=[APQ]+[ADQ]=\dfrac 12PO\cdot AQ+\dfrac 12DO\cdot AQ=\dfrac 12AQ\cdot DP.$ We are done.

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Denote $AM \cap BC = E$, because $\angle MEC = 45^\circ + \angle MAC = 45^\circ + \angle QDC = \angle MQC$, $C,M,E,Q$ are concyclic, so $\angle QEC = \angle QMC = \angle DBC$, hence $EQ\parallel BD$. Then we have ${S_{\Delta QPD}} = {S_{\Delta EPD}} = {S_{\Delta ABP}}$, so ${S_{APQD}} = {S_{ABD}} = \frac{1}{2}{S_{ABCD}}$.

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See that (straight forward) $\triangle PAD\sim\triangle DQA$ hence $\frac{AD}{AQ}=\frac{PD}{AD}\iff AD^2=AQ\cdot PD$, done. Best regards, sunken rock

Did anyone coordinate bash this one? Cause I tried it and it seems to work (although it is painstaking)

I would like to give this a coordinate bash. Choss $D$ to be origin,and $B(2,2)$. The equation of circle $(O)$ is $(x-1)^2+y-1)^2=2$.Let $M(x_0,y_0)$ with $2<x_0\le 3$.After some easy culculation,we get $P(\frac {2x_0}{x_0-y_0+2},\frac {2x_0}{x_0-y_0+2}),Q(\frac {2x_0}{x_0+y_0},\frac {2x_0y_0}{x_0^2+x_0y_0})$. Then $S_{APQD}=S_{APD}+S_{QPD}=\frac {2x_0}{x_0-y_0+2}+\frac {2(x_0-y_0)x_0}{(x_0+y_0)(x_0-y_0+2)}=\frac {4x_0^2}{(x_0+y_0)(x_0-y_0+2)}=\frac {4x_0^2}{x_0^2-y_0^2+2x_0+2y_0}=\frac {4x_0^2}{2x_0^2}=2=\frac {1}{2}S_{ABCD}$. $\blacksquare$

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Miniatures%20Geometriques%20addendum%20IV.pdf p. 59 Sincerely Jean-Louis