In the diagram, $A$, $B$, $C$ and $D$ lie on a circle. $QT$ is a tangent to the circle at $C$. $AT$ is a straight line that intersects the circle at $B$. Chords $AC$ and $BD$ intersect at $P$ and $AB = BC$. Show that

1. Identify equal angles using the alternate segment theorem.

Let $\angle BCT = x$.
By the alternate segment theorem, the angle between the tangent $QT$ and chord $BC$ is equal to the angle in the alternate segment.
Therefore, $\angle BAC = \angle BCT = x$.

2. Use the property of isosceles triangles to show the angles are equal.

Since $AB = BC$ (given), $\triangle ABC$ is an isosceles triangle.
Therefore, the base angles are equal: $\angle BCA = \angle BAC = x$.
Since $\angle BCA = x$ and $\angle BCT = x$, we have $\angle BCA = \angle BCT$.
Thus, the line $BC$ bisects $\angle ACT$.

1. Identify a common angle between the two triangles

In $\triangle ATC$ and $\triangle CTB$,
$\angle ATC = \angle CTB$ (Common angle)

2. Identify a second pair of equal angles using circle properties

$\angle CAT = \angle BCT$ (Alternate segment theorem)

3. Conclude similarity

Since two pairs of corresponding angles are equal, $\triangle ATC$ is similar to $\triangle CTB$ (AA similarity).

1. Express the sum of the two angles using the angle sum of a triangle.

In $\triangle BCT$, the sum of angles is $180^\circ$.
$\angle BCT + \angle BTC = 180^\circ - \angle CBT$

2. Relate $\angle CBT$ to the interior angles of $\triangle ABC$.

Since $A, B, T$ form a straight line, $\angle CBT$ is the exterior angle of $\triangle ABC$.
$\angle CBT = \angle BAC + \angle BCA$
Given $AB = BC$, $\triangle ABC$ is isosceles, so $\angle BAC = \angle BCA$.
Therefore, $\angle CBT = 2 \angle BAC$.

3. Use circle properties to link $\angle BAC$ to $\angle PDC$ and conclude the proof.

Angles subtended by the same arc $BC$ in the same segment are equal, so $\angle BAC = \angle BDC$.
Since $P$ lies on the chord $BD$, $\angle BDC$ is the same as $\angle PDC$.
Thus, $\angle CBT = 2 \angle PDC$.
Substituting this back into the first equation:
$\angle BCT + \angle BTC = 180^\circ - 2 \angle PDC$ (Shown)