Estonian Math Competitions

From the properties of cyclic quadrilaterals we get $\angle KAB = \angle KC'D$ and $\angle KBA = \angle KDC$. Let $A', B', K'$ be the feet of the altitudes of the triangle $ABK$ drawn from the vertices $A, B, K$, respectively, and let $H$ be the orthocenter of the triangle $ABK$ (Fig. 18). The points $A, B, A', B'$ lie on a common circle, hence $\angle KA'B' = \angle KAB$ if $A' \neq B'$. Therefore $A'$ and $B'$ lie on a line parallel to $CD$. Denote this line by $A'B'$ (even in the case $A' = B' = K$).

Fig. 18

Fig. 19

Let $d(X, l)$ be the distance of point $X$ from line $l$, and let $S_\Delta$ be the area of triangle $\Delta$. By two angles, $\angle ACK \sim \angle BDK$ and $\angle PAD \sim \angle PBC$, whence
$$\frac{|AK|}{|BK|} = \frac{|AC|}{|BD|} \text{ and } \frac{|AP|}{|BP|} = \frac{|AD|}{|BC|}. \text{ At the same time}$$
$$\begin{aligned} \frac{d(A, CD)}{d(B, CD)} &= \frac{S_{\triangle ACD}}{S_{\triangle BCD}} = \frac{|AC| \cdot |AD| \cdot \sin \angle CAD}{|BD| \cdot |BC| \cdot \sin \angle CBD} = \frac{|AC| \cdot |AD|}{|BD| \cdot |BC|}, \\ \frac{d(A, KP)}{d(B, KP)} &= \frac{S_{\triangle AKP}}{S_{\triangle BKP}} = \frac{|AK| \cdot |AP| \cdot \sin \angle KAP}{|BK| \cdot |BP| \cdot \sin \angle KBP} = \frac{|AK| \cdot |AP|}{|BK| \cdot |BP|}. \end{aligned}$$
Therefore
$$\frac{|AL|}{|LB|} = \frac{d(A, KP)}{d(B, KP)} = \frac{d(A, CD)}{d(B, CD)}.$$

$$\frac{|AK'|}{|K'B|} = \frac{d(A, KH)}{d(B, KH)} = \frac{d(A, A'B')}{d(B, A'B')}.$$
This equality holds also in the special case $A' = B' = K$. Indeed, let the projections of points $A$ and $B$ to the line $A'B'$ be $X$ and $Y$ correspondingly (Fig. 19), then $\angle AKX = \angle KDC = \angle KBA = \angle AKK'$, $\angle BKY = \angle KCD = \angle KAB = \angle BKK'$, whence $\triangle AKX \sim \triangle AKK'$ and $\triangle BKY \sim \triangle BKK'$. It follows that $|AK'| = |AX|$, $|BK'| = |BY|$ and $\frac{|AK'|}{|K'B|} = \frac{d(A,A'B')}{d(B,A'B')}.$

Considering instead of the cyclic quadrilateral $ABCD$ the quadrilateral determined by points $A, B, A', B'$, and instead of $P$ and $L$ the points $H$ and $K'$ correspondingly, we get similarly that
$$\frac{|AK'|}{|K'B|} = \frac{d(A, KH)}{d(B, KH)} = \frac{d(A, A'B')}{d(B, A'B')}.$$

Since $C$ and $D$ lie on the shorter arc $AB$, we have $\angle BCA = \angle BDA > \frac{\pi}{2}$. Thus the line $A'B'$ is farther from the points $A$ and $B$ than the line $CD$. Since $|AD| > |BC|$, we have $\angle ABD > \angle CAB$ and also $\angle KBA > \angle KAB$, which implies $|KA| > |KB|$. Hence $d(A, CD) + d(K, CD) > d(B, CD) + d(K, CD)$, or $d(A, CD) > d(B, CD)$. All together

Fig. 20

$$\frac{|AL|}{|LB|} = \frac{d(A, CD)}{d(B, CD)} > \frac{d(A, A'B')}{d(B, A'B')} = \frac{|AK'|}{|K'B|}.$$
Hence $L$ lies farther from $A$ than $K'$ on the segment $AB$, therefore $\angle ALK < \angle AK'K = \frac{\pi}{2}$, i.e. $\angle ALK$ is acute.

Denote $\angle KAB = \angle KCD = \alpha$, $\angle KBA = \angle KDC = \beta$, $\angle KAC = \angle KBD = \delta$ and $\angle ALK = \xi$ (Fig. 20). Then $\angle KDB = \alpha + \beta - \delta = \angle KCA$. The condition $|AD| > |BC|$ is equivalent to $\beta > \alpha$, and points $C$ and $D$ being located in the shorter arc $AB$ is equivalent to the inequality $\alpha + \beta - \delta < \frac{\pi}{2}$. In triangle $KCD$ we get $\frac{|KD|}{|KC|} = \frac{\sin \alpha}{\sin \beta}$. From triangles $KDP$ and $KCP$ we obtain $\frac{|KP|}{\sin(\alpha+\beta-\delta)} = \frac{|KD|}{\sin(\xi-(\beta-\delta))}$, $\frac{|KP|}{\sin(\alpha+\beta-\delta)} = \frac{|KC|}{\sin(\xi+(\alpha-\delta))}$, respectively. Consequently, $\frac{|KD|}{|KC|} = \frac{\sin(\xi-(\beta-\delta))}{\sin(\xi+(\alpha-\delta))}$. Expressions of $\frac{|KD|}{|KC|}$ together yield
$$\sin \zeta (\sin \beta \cos(\beta - \delta) - \sin \alpha \cos(\alpha - \delta)) = \cos \zeta (\sin \beta \sin(\beta - \delta) + \sin \alpha \sin(\alpha - \delta)). \quad (1)$$
Clearly $\sin \zeta > 0$ and $\sin \beta \sin(\beta - \delta) + \sin \alpha \sin(\alpha - \delta) > 0$ because $\zeta < \pi$ and $\beta > \delta$, $\alpha > \delta$. By the formula $\sin x \cos y = \frac{1}{2}(\sin(x + y) + \sin(x - y))$, we get $\sin \beta \cos(\beta - \delta) - \sin \alpha \cos(\alpha - \delta) = \frac{1}{2}(\sin(2\beta - \delta) - \sin(2\alpha - \delta))$. Now $\alpha + \beta - \delta < \frac{\pi}{2}$ implies $(2\beta - \delta) + (2\alpha - \delta) < \pi$, i.e., there exists a triangle whose two angles are $2\beta - \delta$ and $2\alpha - \delta$. But $2\beta - \delta > 2\alpha - \delta$ since $\beta > \alpha$, therefore the law of sines in that triangle implies $\sin(2\beta - \delta) > \sin(2\alpha - \delta)$ (the larger the angle, the larger its opposite side in a triangle). Hence the second factor in the l.h.s. of equation (1) is positive. Altogether, we obtain $\cos \zeta > 0$, whence $\zeta < \frac{\pi}{2}$.