In the diagram, A, B C and D lie on a circle, centre O. PQ and PR are tangents to the circle at A and D respectively. It is given that ∠AOC = 140°, ∠DCO = 28° and OA is parallel to CB.

1. Find $\angle BCO$ using the parallel lines $OA \parallel CB$.

Since $OA \parallel CB$, the angle between $CB$ and $CO$ is supplementary to $\angle AOC$.

$\angle BCO = 180^\circ - 140^\circ = 40^\circ$

2. Show that $CA$ divides $\angle BCO$ into two equal angles.

In $\triangle AOC$, $OA = OC$ because both are radii of the circle.

So $\triangle AOC$ is isosceles and

$\angle ACO = \angle OAC = \frac{180^\circ - 140^\circ}{2} = 20^\circ.$

Since $CA$ lies inside $\angle BCO$,

$\angle BCA = \angle BCO - \angle ACO = 40^\circ - 20^\circ = 20^\circ.$

Thus

$\angle BCA = \angle ACO,$

so $CA$ bisects $\angle BCO$.

1. Use part (a) to state the angle at $C$ made by $BC$ and $CA$.

From part (a), $CA$ bisects $\angle BCO$ and $\angle BCA = 20^\circ$.

2. Use the circle theorem that angles in the same segment are equal.

Since $\angle BAC$ and $\angle BCA$ are angles in the same segment, subtended by the same chord $BC$,

$\angle BAC = \angle BCA = 20^\circ$.

1. Find the angle subtended by chord $AD$ at the circumference, $\angle DCA$.

From part (a), $\angle ACO = 20^\circ$.

Given $\angle DCO = 28^\circ$,
$\angle DCA = \angle DCO + \angle OCA = 28^\circ + 20^\circ = 48^\circ.$

2. Use the tangent-chord theorem to find $\angle PDA$.

Since $PD$ is a tangent at $D$ and $DA$ is a chord,
$\angle PDA = \angle DCA$
by the tangent-chord theorem.

Hence
$\angle PDA = 48^\circ.$

1. Use the tangent angle from part (b)(ii) to test whether a regular polygon is possible.

From part (b)(ii), $\angle PDA=48^\circ$.

Since $DP$ and $DR$ are opposite rays on the same tangent line,
$\angle ADR=180^\circ-48^\circ=132^\circ.$

For a regular polygon with interior angle $132^\circ$, the exterior angle would be
$180^\circ-132^\circ=48^\circ.$

Hence
$n=\frac{360^\circ}{48^\circ}=7.5.$

Since $n$ is not a whole number, Lizzy's claim is not possible.