In the figure, AT and CT are tangents to the circle at A and C respectively. AB produced meets TC at P. Given that ∠APC = 60°, ∠ATC = 40° and CD = BC, calculate the following angles and show all the reasons.

1. Identify the equal sides and calculate the base angle in triangle $ACT$.

Since $AT$ and $CT$ are tangents from the same external point $T$, $AT = CT$.

Hence $\triangle ACT$ is isosceles, so
$\angle ACT = \angle CAT.$

Given $\angle ATC = 40^\circ$,
$\angle ACT = \frac{180^\circ - 40^\circ}{2} = 70^\circ.$

1. Find the angle between $AC$ and $AB$ using triangle $APC$.

Since $P$ lies on the tangent $TC$, $\angle ACP = \angle ACT = 70^\circ$.

In $\triangle APC$,
$\angle CAP = 180^\circ - 60^\circ - 70^\circ = 50^\circ.$
Since $A, B, P$ are collinear, $\angle CAP = \angle CAB = 50^\circ$.

2. Use equal chords to find $\angle CAD$.

Given $CD = BC$, equal chords subtend equal angles at the circumference.

So
$\angle CAD = \angle CAB = 50^\circ.$

1. Apply the tangent-chord theorem using chord $AC$ and tangent $CT$.

From part (a),
$\angle ACT = 70^\circ.$
By the tangent-chord theorem, the angle between tangent $CT$ and chord $CA$ is equal to the angle in the alternate segment subtended by chord $CA$.

Hence,
$\angle CBA = \angle ACT = 70^\circ.$

1. Use opposite angles of a cyclic quadrilateral to find $\angle ADC$.

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

Opposite angles of a cyclic quadrilateral are supplementary, so
$\angle ADC + \angle CBA = 180^\circ.$
Using $\angle CBA = 70^\circ$,
$\angle ADC = 180^\circ - 70^\circ = 110^\circ.$