In the diagram, O is the centre of the circle. DOEFG and AEB are straight lines and GA is a tangent to the circle at A. Angle AGD = 32° and angle BCD = 106°.

1. Identify the right angle between the tangent and radius

Since $GA$ is a tangent at $A$ and $OA$ is a radius,
$\angle OAG = 90^\circ.$

2. Use the angle sum of triangle $AOG$ to find $\angle GOA$

Since $D,O,E,F,G$ are collinear, $\angle AGO = \angle AGD = 32^\circ$.

In $\triangle AOG$,
$\angle GOA = 180^\circ - 90^\circ - 32^\circ = 58^\circ.$

1. Identify the right angle subtended by diameter $DF$ at $C$

Since $D, O, F$ are collinear and $O$ is the centre, $DF$ is a diameter of the circle.

Hence,
$\angle DCF = 90^\circ$
(angle in a semicircle).

2. Subtract to find $\angle BCF$

Given
$\angle BCD = 106^\circ,$
and $CF$ lies inside $\angle BCD$.

Therefore,
$\angle BCF = \angle BCD - \angle DCF$
$= 106^\circ - 90^\circ$
$= 16^\circ.$

1. Find the angle between chord $AD$ and radius $AO$

From part (a), $\angle GOA=58^\circ$.

Since $D,O,G$ are collinear,
$\angle AOD=180^\circ-58^\circ=122^\circ.$

In $\triangle AOD$, $OA=OD$ as radii, so
$\angle OAD=\angle ODA=\frac{180^\circ-122^\circ}{2}=29^\circ.$

2. Use the cyclic quadrilateral and tangent-chord theorem to find $\angle BDA$

Since $A,B,C,D$ lie on the circle, opposite angles of cyclic quadrilateral $ABCD$ are supplementary:
$\angle BAD=180^\circ-106^\circ=74^\circ.$

Hence
$\angle BAO=\angle BAD-\angle OAD=74^\circ-29^\circ=45^\circ.$

Since $GA$ is tangent at $A$, $GA\perp AO$, so the angle between tangent $GA$ and chord $AB$ is
$\angle GAB=90^\circ-45^\circ=45^\circ.$

By the tangent-chord theorem,
$\angle BDA=\angle GAB=45^\circ.$

1. Find $\angle DAE$ using the cyclic quadrilateral

Since $A,E,B$ are collinear, $\angle DAE=\angle DAB$.

In cyclic quadrilateral $ABCD$,
$\angle DAB+\angle BCD=180^\circ$
$\angle DAE=\angle DAB=180^\circ-106^\circ=74^\circ.$

2. Find $\angle ADE$ using the radius triangle $AOD$

From part (a), $\angle GOA=58^\circ$.

Since $D,O,G$ are collinear,
$\angle AOD=180^\circ-58^\circ=122^\circ.$

As $OA=OD$, triangle $AOD$ is isosceles, so
$\angle ADO=\frac{180^\circ-122^\circ}{2}=29^\circ.$

Since $D,O,E$ are collinear,
$\angle ADE=29^\circ.$

3. Use the angle sum of triangle $ADE$ to find $\angle DEA$

In triangle $ADE$,
$\angle DEA=180^\circ-74^\circ-29^\circ=77^\circ.$