Angle BCD = angle ADC =

Since chords $AB = BC = CD$ are equal, they subtend equal angles at the circumference.

• $\angle ADB = 10^\circ$ (given). This angle subtends arc AB.
• Angles subtended by arc AB: $\angle ACB$. So, $\angle ACB = 10^\circ$.
• Angles subtended by arc BC: $\angle BAC$ and $\angle BDC$. Since arc BC = arc AB, then $\angle BAC = 10^\circ$ and $\angle BDC = 10^\circ$.
• Angles subtended by arc CD: $\angle CAD$ and $\angle CBD$. Since arc CD = arc AB, then $\angle CAD = 10^\circ$ and $\angle CBD = 10^\circ$.

$\angle ADC$ is the sum of $\angle ADB$ and $\angle BDC$.
From Step 1, we know $\angle ADB = 10^\circ$ and $\angle BDC = 10^\circ$.

$$\angle ADC = \angle ADB + \angle BDC = 10^\circ + 10^\circ = 20^\circ$$

Consider $\triangle ACD$. The sum of angles in a triangle is $180^\circ$.
From Step 1, we know $\angle CAD = 10^\circ$.
From Step 2, we found $\angle ADC = 20^\circ$.

$$\angle ACD = 180^\circ - \angle CAD - \angle ADC$$ $$\angle ACD = 180^\circ - 10^\circ - 20^\circ = 150^\circ$$

The values are:

i) $\boxed{\angle ACD = 150^\circ}$

ii) $\boxed{\angle ADC = 20^\circ}$