图示为一个圆心为 O 的圆，经过 A、B、C、D 和 E 三点。CG 和 FG 分别是该圆在点 C 和 A 处的切线。角 OAB = 56°，角 BDE = 78°，且角 OGC = 19°。求下列各角的值，并给出每一步的理由：

1. Find the angle subtended by chord $AB$ at $E$.

Since $OA=OB$, $\triangle AOB$ is isosceles.

$\angle OBA=\angle OAB=56^\circ$

So
$\angle AOB=180^\circ-56^\circ-56^\circ=68^\circ.$

The angle at the circumference subtended by chord $AB$ is half the angle at the centre:
$\angle AEB=\frac{1}{2}\angle AOB=\frac{1}{2}(68^\circ)=34^\circ.$

2. Find $\angle EAB$ using the cyclic quadrilateral $ABDE$.

Since $A$, $B$, $D$ and $E$ lie on the circle, $ABDE$ is a cyclic quadrilateral.

Opposite angles in a cyclic quadrilateral are supplementary:
$\angle EAB+\angle BDE=180^\circ.$

$\angle EAB=180^\circ-78^\circ=102^\circ.$

3. Use the tangent-chord theorem to find $\angle EAF$.

In $\triangle ABE$,
$\angle EBA=180^\circ-102^\circ-34^\circ=44^\circ.$

By the tangent-chord theorem, the angle between tangent $AF$ and chord $AE$ is equal to the angle in the alternate segment:
$\angle EAF=\angle EBA=44^\circ.$

1. Find the central angle $\angle AOB$ from triangle $AOB$.

Since $OA = OB$ radii, $\triangle AOB$ is isosceles.

Given $\angle OAB = 56^\circ$, so
$\angle ABO = 56^\circ.$
Hence
$\angle AOB = 180^\circ - 56^\circ - 56^\circ = 68^\circ.$

2. Find $\angle AOG$ using the tangent properties.

Since $GC$ and $GA$ are tangents from $G$, the line $OG$ bisects the angle between the two tangents. Therefore
$\angle OGA = \angle OGC = 19^\circ.$
Also, $OA$ is perpendicular to the tangent $GA$, so
$\angle OAG = 90^\circ.$
In $\triangle OAG$,
$\angle AOG = 180^\circ - 90^\circ - 19^\circ = 71^\circ.$

3. Subtract the angles at $O$ to find $\angle BOG$.

From the diagram, $OB$ lies between $OA$ and $OG$.

Therefore
$\angle BOG = \angle AOG - \angle AOB$
$= 71^\circ - 68^\circ = 3^\circ.$