图中显示了一个以 $O$ 为圆心的圆，该圆穿过 $A$、$B$、$C$、$D$ 和 $E$。
$CG$ and $FG$ 分别是圆在点 $C$ 和 $A$ 处的切线。已知角 $OAB = 56^\circ$，角 $BDE = 78^\circ$，角 $OGC = 19^\circ$。 求下列各角的值，并写出每步的理由：

1. Find angle AOB using the isosceles triangle formed by the radii.

Since $OA$ and $OB$ are radii, $OA = OB$, making $\triangle OAB$ an isosceles triangle.
$\angle OBA = \angle OAB = 56^\circ$
$\angle AOB = 180^\circ - 2(56^\circ) = 68^\circ$ (sum of angles in a triangle)

2. Find angle ADB using the angle at centre theorem.

$\angle ADB = \frac{1}{2} \angle AOB = \frac{1}{2}(68^\circ) = 34^\circ$ (angle at centre is twice angle at circumference)

3. Find angle ADE and use the alternate segment theorem to find angle EAF.

$\angle ADE = \angle BDE - \angle ADB = 78^\circ - 34^\circ = 44^\circ$
By the alternate segment theorem, the angle between the tangent $FA$ and chord $AE$ is equal to the angle in the alternate segment.
$\angle EAF = \angle ADE = 44^\circ$

1. Find angle COG using the tangent–radius right angle

Since $CG$ is a tangent and $OC$ is a radius, $OC \perp CG$, so $\angle OCG = 90^\circ$ (tangent $\perp$ radius).

In $\triangle OCG$:
$\angle COG = 180^\circ - 90^\circ - 19^\circ = 71^\circ \quad (\text{angle sum of triangle}).$

2. Use symmetry of the two tangents to find angle AOG

The tangents from external point $G$ to the circle are equal ($GA = GC$), and $\triangle OAG \equiv \triangle OCG$ (RHS). Hence $OG$ bisects $\angle AOC$, so
$\angle AOG = \angle COG = 71^\circ.$

3. Find angle AOB from isosceles triangle OAB

$OA = OB$ (radii), so $\triangle OAB$ is isosceles and $\angle OBA = \angle OAB = 56^\circ$ (base angles of isosceles triangle).
$\angle AOB = 180^\circ - 56^\circ - 56^\circ = 68^\circ.$

4. Combine to find angle BOG

$B$ lies on the arc between $A$ and $C$, with $OG$ inside $\angle AOC$. Since $\angle AOG = 71^\circ$ and $\angle AOB = 68^\circ$:
$\angle BOG = \angle AOG - \angle AOB = 71^\circ - 68^\circ = 3^\circ.$