In the unit circle shown in the figure, chords $PQ$ and $MN$ are parallel to the unit radius $OR$ of the circle with center at $O$. Chords $MP$, $PQ$, and $NR$ are each $s$ units long and chord $MN$ is $d$ units long. 

Of the three equations
\[\textbf{I.}\ d-s=1, \qquad \textbf{II.}\ ds=1, \qquad \textbf{III.}\ d^2-s^2=\sqrt{5}\]
those which are necessarily true are
$\textbf{(A)}\ \textbf{I}\ \text{only} \qquad\textbf{(B)}\ \textbf{II}\ \text{only} \qquad\textbf{(C)}\ \textbf{III}\ \text{only} \qquad\textbf{(D)}\ \textbf{I}\ \text{and}\ \textbf{II}\ \text{only} \qquad\textbf{(E)}\ \textbf{I, II}\ \text{and} \textbf{ III}$
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 $\triangle MOS \cong NOR$ by SAS Congruency.  That means $SM = NR = s$ by CPCTC.

Because $PQNM$ is a [cyclic quadrilateral](https://artofproblemsolving.com/wiki/index.php/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](https://artofproblemsolving.com/wiki/index.php/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, $\triangle POQ \cong \triangle QON \cong \triangle NOR,$ 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](https://artofproblemsolving.com/wiki/index.php/Parallelogram), so $XN = s.$  Thus, $MX = d-s.$

In addition, $PX = s,$ and by the Base Anlge Theorem and [Vertical Angle Theorem](https://artofproblemsolving.com/wiki/index.php?title=Vertical_Angle_Theorem&action=edit&redlink=1), $\angle PMX = \angle MXP = \angle NXR = \angle NRX.$  That means by AAS Congruency, $\triangle MXP = \triangle RXN,$ 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, $\triangle PQY \cong \triangle OYQ.$  That means $PY = OQ = 1.$

Since $\angle PRS$ is an inscribed angle, $\angle POS = 2 \cdot \angle PRS.$  Additionally, $\triangle SOM \cong \triangle MOP$, so $MO$ bisects $\angle SOP.$  Thus, $\angle SOM = \angle PRS,$ making $\triangle PRY \sim \triangle MOS$ by SAS Similarity.

By using the similarity, we find that
\begin{align*} \frac{1}{s} &= \frac{d}{1} \\ ds &= 1 \end{align*}
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{align*} d^2 + 2ds + s^2 &= 5 \\ (d+s)^2 &= 5 \end{align*}
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 $\boxed{\textbf{(E)}}.$