In the diagram, A, B, C and D are points on the circle with centre O. AG and BG are tangents to the circle. GAE, FAD and GBH are straight lines. Angle APD = 71°, angle AGB = 54°.

1. Show that the angles at A and B are right angles

Since $AG$ and $BG$ are tangents to the circle at $A$ and $B$ respectively,
$OA \perp AG \quad \text{and} \quad OB \perp BG.$
Therefore
$\angle OAG=90^\circ \quad \text{and} \quad \angle OBG=90^\circ.$

2. Conclude that A, O, B and G are concyclic and state the centre

Since $\angle OAG$ and $\angle OBG$ are both right angles, A and B lie on the circle with diameter $OG$.

Thus $A,O,B$ and $G$ are concyclic. The centre of this circle is the midpoint of $OG$.

1. Use the angle sum of quadrilateral $AOBG$

Since $AG$ and $BG$ are tangents,
$OA \perp AG$ and $OB \perp BG$.

So $\angle OAG=90^\circ$ and $\angle OBG=90^\circ$.

In quadrilateral $AOBG$,
$\angle AOB+90^\circ+54^\circ+90^\circ=360^\circ$
$\angle AOB=126^\circ$

1. Find the angle at C in triangle $PBC$ using the angle at the centre theorem

From part (b)(i), $\angle AOB=126^\circ$.

Since $\angle ACB$ is the angle at the circumference standing on the same arc $AB$ as the central angle $\angle AOB$,
$\angle ACB=\frac12(126^\circ)=63^\circ.$
Since $P$ lies on $AC$, $\angle PCB=\angle ACB=63^\circ$.

2. Use triangle angle sum in triangle $PBC$

Since chords $AC$ and $BD$ meet at $P$, $\angle CPB$ is vertically opposite to $\angle APD$.
$\angle CPB=71^\circ.$
In triangle $PBC$,
$\angle PBC=180^\circ-71^\circ-63^\circ=46^\circ.$