Estonian Mathematical Olympiad

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 $120^{\circ }$.

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Alternative solution.

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

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Alternative solution.

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