$ABCD$ is a rhombus in which the altitude from $D$ to side $AB$ bisects $AB$. Find the angles of the rhombus.

(A) Let $P$ be the point on $AB$ such that $DP \perp AB$. Given $AP = PB$.
In $\triangle APD$ and $\triangle BPD$:
$AP = BP$ (Given)
$\angle APD = \angle BPD = 90^{\circ}$ (Altitude)
$PD = PD$ (Common side)
By $SAS$ congruence criterion,$\triangle APD \cong \triangle BPD$.
Therefore,$AD = BD$ $(CPCT)$.
Since $ABCD$ is a rhombus,$AD = AB$. Thus,$AD = BD = AB$.
This implies $\triangle ABD$ is an equilateral triangle.
Therefore,$\angle DAB = 60^{\circ}$.
Since opposite angles of a rhombus are equal,$\angle BCD = 60^{\circ}$.
Since consecutive angles are supplementary,$\angle ABC = 180^{\circ} - 60^{\circ} = 120^{\circ}$.
Similarly,$\angle ADC = 120^{\circ}$.
The angles of the rhombus are $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$.