11-th Czech-Slovak-Polish Match, 2011

We use the following lemma

LEMMA. Given two circles ${\Gamma }_1$, ${\Gamma }_2$ with the centre $S_2$ of ${\Gamma }_2$ lying on the circle ${\Gamma }_1$. The circles intersect in two points $K$ and $L$. Let $M$ be the point on the circle ${\Gamma }_1$ (different from $K$ and $L$) and the line $KM$ meets ${\Gamma }_2$ again in $N$. Then $MN=ML$.

Proof. Fig. 1a Fig. 1b For this lemma it is sufficient to prove that the line $M{S}_2$ bisects the angle $\angle NML$. Then in the reflection with respect to the line $M{S}_2$, $ML$ is the image of $MN$. The circle ${\Gamma }_2$ and the point $M$ reflect to themselves, the point $L$ reflects to the point $N$ (the intersection point of $ML$ and ${\Gamma }_2$). So the triangle $△MLN$ is isosceles (some considerations are needed according to the position of the point $M$).

Firstly, let $M$ lie on the arc $\widehat{KL}$ not containing the point $S_2$. As $S_{2}K=S_{2}L$, we directly have $\angle KM{S}_2=\angle S_{2}ML$. Secondly, let $M$ lie on the arc $\widehat{KL}$ containing point $S_2$. Let $R$ be an arbitrary point on the arc $\widehat{KL}$ not containing the point $S_2$. Similarly as before $\angle KR{S}_2=\angle S_{2}RL$, then using identical angles in the cyclic quadrilaterals $R{S}_{2}MK$ and $RL{S}_{2}M$ we obtain $\angle NM{S}_2=\angle KR{S}_2=\angle S_{2}RL=\angle S_{2}ML$. $■$

Let the perpendicular line from the point $E$ to the line $AB$ intersect the arc $\widehat{BC}$ in the point $S$, $k_1$ be the circle centered in $A$ passing through $C$, $k_2$ be the circle centered in $B$ passing through $D$, $k$ be the circle passing through $A$, $B$, $C$, $D$. The line $SC$ intersects $k_1$ again in $C^{\prime }$ and the line $SD$ intersects $k_2$ again in $D^{\prime }$. Circles $k_1$ and $k$ meet in $C$, $C^{\prime \prime }$ and circles $k_2$ and $k$ meet in $D$, $D^{\prime \prime }$. Using the lemma we have $S{C}^{\prime }=S{C}^{\prime \prime }$ and $S{D}^{\prime }=S{D}^{\prime \prime }$. Let the circles $k_1$ and $k_2$ meet again in $E^{\prime }$. Using a contradiction, we shall prove that $C^{\prime \prime }=D^{\prime \prime }=E^{\prime }$.

The point $S$ lies on the chord $E{E}^{\prime }\perp AB$ of the circles $k_1$ and $k_2$. That means its powers to these two circles are equal. We know that $S$ bisects the arc $\widehat{CD}$, so $SD=SC$ and consequently $S{C}^{\prime \prime }=S{C}^{\prime }=S{D}^{\prime }=S{D}^{\prime \prime }$. If $D^{\prime \prime }$ and $C^{\prime \prime }$ are different points then the triangle $△S{D}^{\prime \prime }{C}^{\prime \prime }$ is isosceles and its altitude from $S$ passes through the circumcentre of $k$ and consequently (by the symmetry) the quadrilateral $CD{D}^{\prime \prime }{C}^{\prime \prime }$ is isosceles trapezoid ($SC=SD$). We can find points $A$ and $B$ as the intersection points of the axes of the segments $C{C}^{\prime \prime }$ and $D{D}^{\prime \prime }$ with $k$. But then also $ABCD$ is an isosceles trapezoid, $AB\parallel CD$, which is a contradiction to the given $AB\nparallel CD$.

If we denote $\angle D{E}^{\prime }S=\angle S{E}^{\prime }C=\alpha$ and $\angle A{E}^{\prime }D=\beta$ then $\angle C{E}^{\prime }D=2\alpha -\beta$ because $2|\widehat{CD}|=|\widehat{AB}|$. Using $BD=B{E}^{\prime }$ and $AC=A{E}^{\prime }$ we compute the angles in the triangle $AB{E}^{\prime }$: $$\begin{array}{rl}2\alpha +\beta & =\angle A{E}^{\prime }C=\angle AC{E}^{\prime }=\angle AB{E}^{\prime }\\ 4\alpha -\beta & =\angle B{E}^{\prime }D=\angle BD{E}^{\prime }=\angle BA{E}^{\prime }\end{array}$$ which yields to $$180^{\circ }=2\alpha +\beta +4\alpha -\beta +4\alpha =10\alpha$$ and $\angle ACB=180^{\circ }-\angle A{E}^{\prime }B=180^{\circ }-4\alpha =108^{\circ }$.