smc

First, let $S$ be on circle $O$ so $RS$ is a diameter. In order to prove that the three statements are true (or false), we first show that $SM=QN=s,$ and then we examine each statement one by one. Lemma 1: $SM=QN=s$ Since $OM=ON,$ by the Base Angle Theorem, $\angle OMN=\angle ONM.$ By the Alternate Interior Angles Theorem, $\angle MOS=\angle NMO=\angle MNO=\angle NOR,$ making $△MOS\cong NOR$ by SAS Congruency. That means $SM=NR=s$ by CPCTC. Because $PQNM$ is a cyclic quadrilateral, $\angle MPQ+\angle QNM=180^{\circ }$, but we are given that $PQ$ is parallel to $MN$, so $\angle MPQ+\angle PMN=180^{\circ }$. Therefore, $\angle PMN=\angle QNM.$ That makes $PQNM$ an isosceles trapezoid, so $MP=QN=s.$ Lemma 2: Showing Statement I is (or isn't) true Let $X$ be the intersection of $MN$ and $PR.$ By SSS Congruency, $△POQ\cong △QON\cong △NOR,$ so $\angle PQN=\angle QNR.$ We know that $PQNR$ is a cyclic quadrilateral, so $\angle PQN+\angle PRN=\angle QNR+\angle PRN=180^{\circ },$ so $QN\parallel PR.$ That makes $PQNX$ a parallelogram, so $XN=s.$ Thus, $MX=d-s.$ In addition, $PX=s,$ and by the Base Anlge Theorem and Vertical Angle Theorem, $\angle PMX=\angle MXP=\angle NXR=\angle NRX.$ That means by AAS Congruency, $△MXP=△RXN,$ so $MX=XR=d-s$. By the Base Angle Theorem and the Alternating Interior Angle Theorem, $\angle XRM=\angle XMR=\angle MRO=\angle RMO,$ so by ASA Congruency, $\angle MXR=\angle MOR.$ Thus, $MX=OR=d-s=1.$ Statement I is true. Lemma 3: Showing Statement II is (or isn't) true From Lemma 2, we have $PX+XR=s+(d-s)=d$. Draw point $Y$ on $SR$ such that $YR=d,$ making $YO=s.$ Since $PQ\parallel YO,$ we have $\angle QPO=\angle POY.$ Additionally, $PQ=YO$ and $PO=PO,$ so by SAS Congruency, $△PQY\cong △OYQ.$ That means $PY=OQ=1.$ Since $\angle PRS$ is an inscribed angle, $\angle POS=2\cdot \angle PRS.$ Additionally, $△SOM\cong △MOP$, so $MO$ bisects $\angle SOP.$ Thus, $\angle SOM=\angle PRS,$ making $△PRY\sim △MOS$ by SAS Similarity. By using the similarity, we find that $$\begin{array}{rl}\frac{1}{s}& =\frac{d}{1}\\ ds& =1\end{array}$$ Thus, Statement II is true. Lemma 4: Showing Statement III is (or isn't) true From Lemmas 2 and 3, we have $d-s=1$ and $ds=1$. Squaring the first equation results in $$d^2-2ds+s^2=1.$$ Adding $4ds=4$ to both sides results in $$\begin{array}{rl}{d}^2+2ds+s^2& =5\\ (d+s{)}^2& =5\end{array}$$ Since $d+s$ is positive, we find that $d+s=\sqrt{5},$ which confirms that Statement III is true. In summary, all three statements are true, so the answer is $\boxed{\text{I, II}\text{and}\text{ III}}.$