In the diagram, ABCD is a rectangle. OAB is a sector of a circle, centre O. OXD and OYC are straight lines.

1. State the first pair of equal sides.

$OA=OB$

Reason: $OA$ and $OB$ are radii of the same circle, centre $O$.

2. State the second pair of equal sides.

$AD=BC$

Reason: opposite sides of a rectangle are equal.

3. Show the included angles are equal and conclude congruence.

Since $OA=OB$, $\triangle OAB$ is isosceles.

Therefore,
$\angle OAB=\angle ABO$.

Also, since $ABCD$ is a rectangle,
$\angle DAB=\angle ABC=90^\circ$.

Hence
$\angle OAD=90^\circ-\angle OAB$
and
$\angle OBC=90^\circ-\angle ABO.$

Since $\angle OAB=\angle ABO$,
$\angle OAD=\angle OBC.$

Therefore, $\triangle OAD \cong \triangle OBC$ by SAS.

1. Establish the equal corresponding side ratios.

From part (a), $\triangle OAD \cong \triangle OBC$, so $OD=OC$.

Also, $OX=OY$ since $X$ and $Y$ lie on the same circular arc with centre $O$.

Therefore,
$\frac{OX}{OD}=\frac{OY}{OC}.$

2. Show the included angle is equal and conclude similarity.

Since $O, X, D$ are collinear and $O, Y, C$ are collinear,
$\angle XOY=\angle DOC.$

Hence, $\triangle OXY \sim \triangle ODC$ by SAS similarity.