B、C、D、E 和 F 是以 O 为圆心的圆上的点。 AE 是圆的切线，ABC 是一条直线。 $\angle BAE = 38^\circ$、$\angle BOE = 72^\circ$ 且 $\angle DCE = 40^\circ$。 求以下角的大小，并写出每一步的理由：

1. Find the angle between the radius and the tangent at $E$.

Since $AE$ is tangent to the circle at $E$ and $OE$ is a radius,
$\angle OEA = 90^\circ.$

2. Use the angle sum of the quadrilateral path via triangle angles to find $\angle OBA$.

In triangle $AOE$,
$\angle OAE = \angle BAE = 38^\circ,$
so
$\angle AOE = 180^\circ - 90^\circ - 38^\circ = 52^\circ.$

Since $\angle BOE = 72^\circ$,
$\angle AOB = 72^\circ - 52^\circ = 20^\circ.$

In triangle $AOB$, $OA = OB$ is not needed; instead, since $A,B,C$ are collinear and $AOB$ is formed with $A,O,B$, the required angle is the angle between $OB$ and $BA$. From the diagram, $\angle OBA = 20^\circ$.

1. Find $\angle ECB$ using the angle at the centre theorem.

Since $\angle BOE = 72^\circ$, the angle at the circumference subtended by the same chord $BE$ is

$\angle ECB = \dfrac{1}{2}\times 72^\circ = 36^\circ$.

Reason: angle at the centre is twice the angle at the circumference.

2. Find $\angle DCB$ and use opposite angles in a cyclic quadrilateral.

At $C$,

$\angle DCB = \angle DCE + \angle ECB$

$= 40^\circ + 36^\circ = 76^\circ$.

Since $B, C, D, E$ lie on the circle, opposite angles of cyclic quadrilateral $BCDE$ are supplementary:

$\angle DEB = 180^\circ - 76^\circ = 104^\circ$.

1. Find the angle subtended by chord $BE$ at $C$.

Since $\angle BOE = 72^\circ$, the angle at the circumference standing on the same arc $BE$ is
$\angle ECB = \frac{1}{2}\times 72^\circ = 36^\circ.$

2. Find the angle between $CE$ and the tangent $AE$.

Since $A$, $B$ and $C$ are on a straight line,
$\angle ECA = \angle ECB = 36^\circ.$
In triangle $AEC$,
$\angle AEC = 180^\circ - 38^\circ - 36^\circ = 106^\circ.$

3. Use the radius-tangent property to find $\angle OEC$.

Since $AE$ is tangent to the circle at $E$, $OE \perp AE$.
So
$\angle OEA = 90^\circ.$
Therefore
$\angle OEC = \angle AEC - \angle AEO = 106^\circ - 90^\circ = 16^\circ.$