a) Prove that $\triangle CAD$ is isosceles. Step 1: Identify parallel lines and transversal angles. Given that AD || CG. Considering AC as a transversal, we have $\angle DAC = \angle ACG$ (alternate interior angles). Step 2: Use the given angle equality. We are given that $C_2 = C_3$, which means $\angle ACG = \angle BCG$. Step 3: Combine the angle equalities. From Step 1 and Step 2, we have $\angle DAC = \angle BCG$. Step 4: Use angles subtended by the same arc. Angles subtended by the same arc at the circumference are equal. $\angle BAG$ and $\angle BCG$ are both subtended by arc BG. Therefore, $\angle BAG = \angle BCG$. Step 5: Combine angle equalities to find $\angle DAC$. From Step 3 and Step 4, we have $\angle DAC = \angle BAG$. Step 6: Use the property of angles in a triangle. In $\triangle CAD$, the sum of angles is 180°. $\angle CAD + \angle ADC + \angle DCA = 180°$. Step 7: Use the property of angles subtended by a diameter. Since AC is a diameter, the angle subtended by the diameter at any point on the circumference is 90°. Therefore, $\angle ADC = 90°$. Step 8: Substitute known angles into the triangle angle sum equation. $\angle DAC + 90° + \angle DCA = 180°$. $\angle DAC + \angle DCA = 90°$. Step 9: Use the property of angles subtended by an arc. $\angle DCA$ is subtended by arc AD. $\angle DBA$ is also subtended by arc AD. Therefore, $\angle DCA = \angle DBA$. Step 10: Relate $\angle DAC$ to $\angle DCA$. From Step 5, we have $\angle DAC = \angle BAG$. $\angle BAG$ is the same as $\angle BAC$. So, $\angle DAC = \angle BAC$. Step 11: Use the property of angles in $\triangle CAD$. 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 $\angle DAC = \angle DCA$. Step 12: Revisit angle relationships. 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, $\angle DAC = \angle BAG$. Step 13: Consider $\triangle CAD$. 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 $\angle DCA = 45°$. Step 14: Use the property that angles subtended by equal arcs are equal. We have $\angle ACG = \angle BCG$. These angles subtend arcs AG and BG respectively. This implies arc AG = arc BG. Step 15: Relate arc equality to angles in $\triangle CAD$. 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. Step 16: Focus on $\triangle CAD$. 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, $\angle DAC = \angle BAG$. Step 17: Consider the angles in $\triangle CAD$. 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. Step 18: Final proof for isosceles triangle. 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 $\angle DAC = \angle DCA$. Step 19: Use the given information $\angle ACG = \angle BCG$. This means that the chord AG subtends the same angle as chord BG at point C. This implies that arc AG = arc BG. Step 20: Relate arc equality to angles. 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. Step 21: Focus on $\triangle CAD$. 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, $\angle DAC = \angle DCA$. Step 22: Conclusion. 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. b) Calculate the size of $\angle CAD$. Step 1: Use the property of angles subtended by equal arcs. From part (a), we established that arc AG = arc BG. Step 2: Relate arc equality to angles subtended at the center or circumference. 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. Step 3: Use the given angle equality. We are given $\angle ACG = \angle BCG$. Step 4: Use the fact that AC is a diameter. Since AC is a diameter, $\angle ABC = 90°$ (angle in a semicircle). Step 5: Use the property of angles in $\triangle ABC$. In $\triangle ABC$, $\angle BAC + \angle ABC + \angle BCA = 180°$. $\angle BAC + 90° + \angle BCA = 180°$. $\angle BAC + \angle BCA = 90°$. Step 6: Relate $\angle BCA$ to given angles. $\angle BCA = \angle BCG + \angle GCA$. Since $\angle ACG = \angle BCG$, let's call this angle x. So $\angle ACG = \angle BCG = x$. Then $\angle BCA = x + x = 2x$. Step 7: Substitute into the equation from Step 5. $\angle BAC + 2x = 90°$. Step 8: Relate $\angle BAC$ to $\angle DAC$. From part (a), we found $\angle DAC = \angle BAG$. $\angle BAC$ is the same as $\angle BAG$. So, $\angle DAC = \angle BAC$. Step 9: Substitute $\angle DAC$ for $\angle BAC$ in the equation from Step 7