The problem states that $ABCD$ is a cyclic quadrilateral with $AB = AD$ and $BC = DC$. $AC$ is the diameter of the circle, and $\angle ADB = 10^\circ$. (a) We need to determine the special name given to cyclic quadrilateral $ABCD$. (b) We need to find the measures of $\angle ACD$ and $\angle ADC$.

1. Problem Description

The problem states that $ABCD$ is a cyclic quadrilateral with $AB = AD$ and $BC = DC$. $AC$ is the diameter of the circle, and $\angle ADB = 10^\circ$.

(a) We need to determine the special name given to cyclic quadrilateral $ABCD$.

(b) We need to find the measures of $\angle ACD$ and $\angle ADC$.

2. Solution Steps

(a) Since $AB = AD$ and $BC = DC$, the quadrilateral $ABCD$ has two pairs of adjacent sides that are equal in length. This makes it a kite. Since $ABCD$ is a cyclic quadrilateral, it is a cyclic kite. Since $AC$ is the diameter, it cuts the kite into two right triangles so $ABCD$ is a special kind of cyclic kite.

(b) (i) We are given that $\angle ADB = 10^\circ$. Since angles subtended by the same chord at the circumference are equal, $\angle ACB = \angle ADB = 10^\circ$.

Also, since $AC$ is the diameter, $\angle ADC = 90^\circ$ and $\angle ABC = 90^\circ$. In $\triangle ADC$, we have $\angle DAC + \angle ACD + \angle ADC = 180^\circ$.

Since $AD=AB$, $\angle ADB = \angle ACB = 10^{\circ}$. Since $AC$ is a diameter, $\angle ABC = 90^{\circ}$ and $\angle ADC = 90^{\circ}$. $\angle ACD$ is the angle we are seeking.

Since $\angle ADC = 90^\circ$, $\angle DAC = \angle BAC = 90^\circ - \angle ACD$ but we don't know $\angle ACD$.

Since $\angle ACB = 10^{\circ}$ and $\angle ABC = 90^{\circ}$, $\angle BAC = 180^{\circ} - (90^{\circ} + 10^{\circ}) = 80^{\circ}$.

Since $AB=AD$, $\triangle ABD$ is an isosceles triangle, so $\angle ABD = \angle ADB = 10^{\circ}$, hence $\angle DAB = 180^{\circ} - 2(10^{\circ}) = 160^{\circ}$.

Also $\angle DAC = \angle DAB - \angle CAB$. $\angle CAB = 80^\circ$. Hence $\angle DAC = 160^\circ - 80^\circ = 80^\circ$.

In $\triangle ADC$, $\angle DAC + \angle ACD + \angle ADC = 180^\circ$, so $80^\circ + \angle ACD + 90^\circ = 180^\circ$, which implies $\angle ACD = 180^\circ - 80^\circ - 90^\circ = 10^\circ$.

(ii) Since $AC$ is a diameter, $\angle ADC = 90^\circ$.

3. Final Answer

(a) Kite

(b) (i) $\angle ACD = 10^\circ$

(ii) $\angle ADC = 90^\circ$