In the diagram, O is the centre of the circle through A, B, C, D and TB is the tangent at B. AC is the diameter of the circle, and the length of the minor arc BAD is 11 cm. Given that ∠OBC = 40°, ∠ACD = 30° and ∠BTC = 70°.

1. Identify equal angles standing on the same chord $AD$.

$\angle ABD = \angle ACD$ because angles in the same segment standing on chord $AD$ are equal.

Given $\angle ACD = 30^\circ$, so
$\angle ABD = 30^\circ.$

1. Find $\angle BCD$ by adding the two angles at $C$.

Since $A$, $O$ and $C$ lie on the diameter $AC$, $CA$ is the same straight line as $CO$.

Therefore
$\angle BCA = \angle BCO = 40^\circ$.

Given $\angle ACD = 30^\circ$,
$\angle BCD = \angle BCA + \angle ACD = 40^\circ + 30^\circ = 70^\circ.$

2. Use opposite angles in the cyclic quadrilateral $ABCD$.

Since $A$, $B$, $C$ and $D$ lie on the same circle, $ABCD$ is a cyclic quadrilateral.

Opposite angles of a cyclic quadrilateral are supplementary, so
$\angle BAD + \angle BCD = 180^\circ.$

Hence
$\angle BAD = 180^\circ - 70^\circ = 110^\circ.$

1. Find the central angle subtended by the minor arc $BAD$.

From 5(a)(ii), $\angle BAD=110^\circ$.

The angle at the centre subtended by the same arc $BD$ is
$\angle BOD=2\times 110^\circ=220^\circ.$
This is the reflex angle, so it corresponds to the minor arc $BAD$.

2. Use the arc length formula to calculate the radius.

Arc length $BAD=11$ cm and $\pi=\frac{22}{7}$.

$11=\frac{220}{360}\times 2\times \frac{22}{7}\times r$
$11=\frac{11}{18}\times \frac{44}{7}r$
$11=\frac{242}{63}r$
$r=11\times \frac{63}{242}=\frac{7}{2}=3.5$

1. Show one pair of corresponding angles is equal.

From part (a)(ii),
$\angle BAD = 110^\circ.$
Since $ABCD$ is a cyclic quadrilateral,
$\angle BCD = 180^\circ - 110^\circ = 70^\circ.$
Also, $\angle CTB = 70^\circ$ is given.
Hence,
$\angle BCD = \angle CTB.$

2. Show a second pair of corresponding angles is equal.

Since $TB$ is tangent at $B$, the radius $OB$ is perpendicular to $TB$.
$\angle OBT = 90^\circ.$
Given $\angle OBC = 40^\circ$,
$\angle CBT = 90^\circ - 40^\circ = 50^\circ.$
Using the angle sum of $\triangle BCD$,
$\angle BDC = 180^\circ - \angle BCD - \angle CBD.$
From part (a)(i), $\angle ABD = 30^\circ$, so
$\angle CBD = \angle CBO + \angle OBA?$
This route requires care, so instead use $\angle CTB=70^\circ$ and the tangent angle found above:
$\angle BCT = 180^\circ - 70^\circ - 50^\circ = 60^\circ.$
Since $A,O,C$ are collinear and $AC$ is a diameter, $\angle ABC=90^\circ$.
Thus,
$\angle CBD = \angle ABC - \angle ABD = 90^\circ - 30^\circ = 60^\circ.$
Hence,
$\angle CBD = \angle BCT.$

3. Conclude the triangles are similar.

Since two pairs of corresponding angles are equal,
$\triangle BCD \sim \triangle CTB \quad \text{(AA similarity).}$
The correspondence is
$B \leftrightarrow C,\quad C \leftrightarrow T,\quad D \leftrightarrow B.$

1. Identify the corresponding sides from the similarity statement.

From part (b)(i),
$\triangle BCD \sim \triangle CTB.$
So the corresponding vertices are
$B \leftrightarrow C,\quad C \leftrightarrow T,\quad D \leftrightarrow B.$
Hence,
$BC \leftrightarrow CT,\quad BD \leftrightarrow CB.$

2. Use proportional corresponding sides to express $CT$ in terms of $x$ and $y$.

Since corresponding sides of similar triangles are proportional,
$\frac{CT}{BC}=\frac{CB}{BD}.$
Given $BC=x$ and $BD=y$,
$\frac{CT}{x}=\frac{x}{y}.$
Therefore,
$CT=\frac{x^2}{y}.$