Let $\angle ABC = \angle ADC = \alpha$; then $\angle AKL = \alpha$ (Fig. 28).

According to Thales' theorem $AK$ is perpendicular to $BC$ and $AL$ is perpendicular to $CD$. But $\angle ABK = \alpha = \angle ADL$ and $AB = AD$, so $ABK$ and $ADL$ are equal. Therefore, $AK = AL$ from which $\angle ALK = \alpha$.

As the sum of the internal angles of a quadrilateral is $360^\circ$, we have
$$\angle KAL = 360^\circ - \angle KCL - \angle AKC - \angle ALC = 360^\circ - (180^\circ - \alpha) - 2 \cdot 90^\circ = \alpha.$$

Fig. 28

Therefore, the triangle $AKL$ is equilateral as all its angles are equal to $\alpha$, implying that the angles of the rhombus are $60^\circ$ and $120^\circ$.

As in Solution 1, denote $\alpha = \angle ABC = \angle ADC = \angle AKL$ and show that triangles $ABK$ and $ADL$ are equal. Therefore, $BK = DL$ from which $CK = CL$. We have $\angle KCL = 180^\circ - \alpha$, therefore $\angle CKL = \angle CLK = \frac{\alpha}{2}$.

Thus $90^\circ = \angle AKC = \angle AKL + \angle LKC = \frac{3}{2}\alpha$ from which $\alpha = 60^\circ$. Therefore, the angles of the rhombus are $60^\circ$ and $120^\circ$.

Fig. 29

As in Solution 1, denote $\alpha = \angle ABC = \angle ADC = \angle AKL$ and show that triangles $ABK$ and $ADL$ are equal. Hence, $BK = DL$ from which $\frac{BK}{BC} = \frac{DL}{CD}$. Thus $KL \parallel BD$ and $AC \perp KL$. Let $AC$ and $KL$ meet at $F$ (Fig. 29).

The altitude $KF$ from the right angle of triangle $CKA$ divides the triangle to two triangles $CFK$ and $KFA$, both similar to $CKA$. Therefore, $\angle ACK = \angle AKF = \alpha$. The diagonal of the rhombus is also the angle bisector, therefore, $\angle BCD = 2\alpha$. Equation $180^\circ - \alpha = 2\alpha$ gives us $\alpha = 60^\circ$ and the angles of the rhombus are $60^\circ$ and $120^\circ$.

Fig. 29