In the figure, circle ABCF and circle CDEF intersect at C and F. AFD and BFE are straight lines and BC is parallel to AD. Angle ACB = $30^\circ$, angle BAF = $100^\circ$ and angle FED = $120^\circ$.

1. Find $\angle BCF$ using the cyclic quadrilateral $ABCF$.

Since $A, B, C, F$ lie on the same circle, opposite angles of cyclic quadrilateral $ABCF$ are supplementary.

$\angle BAF + \angle BCF = 180^\circ$

$100^\circ + \angle BCF = 180^\circ$

$\angle BCF = 80^\circ$

2. Find $\angle ACF$ by subtracting the given angle $\angle ACB$.

At $C$, $\angle BCF$ is made up of $\angle BCA$ and $\angle ACF$.

Given $\angle ACB = 30^\circ$,

$\angle ACF = \angle BCF - \angle ACB$

$\angle ACF = 80^\circ - 30^\circ = 50^\circ$

3. Find $\angle FCD$ using cyclic quadrilateral $CDEF$, then obtain $\angle ACD$.

Since $C, D, E, F$ lie on the same circle, opposite angles of cyclic quadrilateral $CDEF$ are supplementary.

$\angle FED + \angle FCD = 180^\circ$

$120^\circ + \angle FCD = 180^\circ$

$\angle FCD = 60^\circ$

Therefore,

$\angle ACD = \angle ACF + \angle FCD$

$\angle ACD = 50^\circ + 60^\circ = 110^\circ$

1. Find \(\angle CAD\) using \(BC \parallel AD\).

Since $BC \parallel AD$ and $AC$ is a transversal,
$\angle CAD = \angle ACB = 30^\circ$
(alternate angles).

2. Use the angle sum of triangle \(ACD\) to find \(\angle CDA\).

From part (i)(a),
$\angle ACD = 110^\circ.$
In $\triangle ACD$,
$\angle CDA = 180^\circ - 110^\circ - 30^\circ = 40^\circ.$

1. Show that one pair of corresponding angles is equal.

Since $A,G,C$ are collinear and $B,G,F$ are collinear,
$\angle BAG=\angle BAC$
and
$\angle CFG=\angle CFB.$
In circle $ABCF$, angles in the same segment subtended by chord $BC$ are equal, so
$\angle BAC=\angle CFB.$
Hence,
$\angle BAG=\angle CFG.$

2. Show that a second pair of corresponding angles is equal.

Since $B,G,F$ are collinear and $A,G,C$ are collinear,
$\angle ABG=\angle ABF$
and
$\angle FCG=\angle FCA.$
In circle $ABCF$, angles in the same segment subtended by chord $AF$ are equal, so
$\angle ABF=\angle FCA.$
Hence,
$\angle ABG=\angle FCG.$

3. Show that the corresponding included side is equal and conclude congruence.

From part (i)(b), $\angle CAD=30^\circ$. Since $A,F,D$ are collinear,
$\angle CAF=\angle CAD=30^\circ.$
Also given,
$\angle ACB=30^\circ.$
Thus
$\angle ACB=\angle CAF.$
In circle $ABCF$, equal angles subtend equal chords, so
$AB=CF.$
Therefore, with
$\angle BAG=\angle CFG,\quad \angle ABG=\angle FCG,\quad AB=CF,$
we have
$\triangle BAG \cong \triangle CFG \quad \text{(ASA)}.$