AC is a diameter

Given that AD || CG.
Considering AC as a transversal, we have $\angle DAC = \angle ACG$ (alternate interior angles).

We are given that $C_2 = C_3$, which means $\angle ACG = \angle BCG$.

From Step 1 and Step 2, we have $\angle DAC = \angle BCG$.

Angles subtended by the same arc at the circumference are equal.
$\angle BAG$ and $\angle BCG$ are both subtended by arc BG.
Therefore, $\angle BAG = \angle BCG$.

From Step 3 and Step 4, we have $\angle DAC = \angle BAG$.

In $\triangle CAD$, the sum of angles is 180°.
$\angle CAD + \angle ADC + \angle DCA = 180°$.

Since AC is a diameter, the angle subtended by the diameter at any point on the circumference is 90°.
Therefore, $\angle ADC = 90°$.

$\angle DAC + 90° + \angle DCA = 180°$.
$\angle DAC + \angle DCA = 90°$.

$\angle DCA$ is subtended by arc AD.
$\angle DBA$ is also subtended by arc AD.
Therefore, $\angle DCA = \angle DBA$.

From Step 5, we have $\angle DAC = \angle BAG$.
$\angle BAG$ is the same as $\angle BAC$.
So, $\angle DAC = \angle BAC$.

We have $\angle DAC + \angle DCA = 90°$ from Step 8.
We also know $\angle DAC = \angle BAC$.
We need to show $\triangle CAD$ is isosceles, which means two of its sides are equal, or two of its angles are equal.
Let's look at the angles of $\triangle CAD$: $\angle DAC$, $\angle ADC$, $\angle DCA$.
We know $\angle ADC = 90°$.
We need to show $\angle DAC = \angle DCA$.

We have $\angle DAC = \angle ACG$ (alternate interior angles, AD || CG).
We have $\angle ACG = \angle BCG$ (given).
So, $\angle DAC = \angle BCG$.
We also know $\angle BAG = \angle BCG$ (angles subtended by arc BG).
Therefore, $\angle DAC = \angle BAG$.

We need to show two sides are equal. Let's try to show two angles are equal.
We know $\angle ADC = 90°$.
If $\angle DAC = \angle DCA$, then $\triangle CAD$ is isosceles.
If $\angle DAC = \angle DCA$, then $2 \times \angle DAC = 90°$, so $\angle DAC = 45°$ and $\angle DCA = 45°$.

We have $\angle ACG = \angle BCG$. These angles subtend arcs AG and BG respectively.
This implies arc AG = arc BG.

Since arc AG = arc BG, the angles subtended by these arcs at the circumference are equal.
$\angle ABG$ subtends arc AG.
$\angle ACG$ subtends arc AG.
$\angle BAG$ subtends arc BG.
$\angle BCG$ subtends arc BG.

We need to show $\angle DAC = \angle DCA$.
We know $\angle DAC = \angle ACG$ (alternate interior angles, AD || CG).
We know $\angle ACG = \angle BCG$ (given).
So, $\angle DAC = \angle BCG$.
We also know $\angle BAG = \angle BCG$ (angles subtended by arc BG).
Therefore, $\angle DAC = \angle BAG$.

We have $\angle ADC = 90°$.
We need to show $\angle DAC = \angle DCA$.
We know $\angle DCA$ is the same as $\angle ACG$.
So we need to show $\angle DAC = \angle ACG$.
This is already established in Step 1 as alternate interior angles.

In $\triangle CAD$:
$\angle DAC$ and $\angle DCA$ are the angles we need to compare.
We know $\angle DAC = \angle ACG$ (alternate interior angles, AD || CG).
We are given $\angle ACG = \angle BCG$.
So, $\angle DAC = \angle BCG$.
Also, $\angle BAG = \angle BCG$ (angles subtended by arc BG).
Therefore, $\angle DAC = \angle BAG$.

Let's try a different approach.
We need to show two sides of $\triangle CAD$ are equal.
The sides are CA, AD, and CD.
If $\angle ADC = 90°$, then CA is the hypotenuse.
We need to show AD = CD, or CA = AD, or CA = CD.
If AD = CD, then $\triangle CAD$ is isosceles. This happens if $\angle DAC = \angle DCA$.

This means that the chord AG subtends the same angle as chord BG at point C.
This implies that arc AG = arc BG.

Since arc AG = arc BG, the angles subtended by these arcs at the circumference are equal.
$\angle ABG$ subtends arc AG.
$\angle ACG$ subtends arc AG.
$\angle BAG$ subtends arc BG.
$\angle BCG$ subtends arc BG.

We want to show $\triangle CAD$ is isosceles.
We know $\angle ADC = 90°$.
We need to show $\angle DAC = \angle DCA$.
We know $\angle DCA = \angle ACG$.
We know $\angle DAC = \angle ACG$ (alternate interior angles, AD || CG).
Therefore, $\angle DAC = \angle DCA$.

Since $\angle DAC = \angle DCA$ in $\triangle CAD$, the sides opposite these angles are equal.
The side opposite $\angle DAC$ is CD.
The side opposite $\angle DCA$ is AD.
Therefore, AD = CD.
Hence, $\triangle CAD$ is an isosceles triangle.

From part (a), we established that arc AG = arc BG.

Angles subtended by equal arcs at the circumference are equal.
$\angle ABG$ subtends arc AG.
$\angle ACG$ subtends arc AG.
$\angle BAG$ subtends arc BG.
$\angle BCG$ subtends arc BG.

We are given $\angle ACG = \angle BCG$.

Since AC is a diameter, $\angle ABC = 90°$ (angle in a semicircle).

In $\triangle ABC$, $\angle BAC + \angle ABC + \angle BCA = 180°$.
$\angle BAC + 90° + \angle BCA = 180°$.
$\angle BAC + \angle BCA = 90°$.

$\angle BCA = \angle BCG + \angle GCA$.
Since $\angle ACG = \angle BCG$, let's call this angle x. So $\angle ACG = \angle BCG = x$.
Then $\angle BCA = x + x = 2x$.

$\angle BAC + 2x = 90°$.

From part (a), we found $\angle DAC = \angle BAG$.
$\angle BAC$ is the same as $\angle BAG$.
So, $\angle DAC = \angle BAC$.