, 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 $MS_2$ bisects the angle $\angle NML$. Then in the reflection with respect to the line $MS_2$, $ML$ 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 $\triangle MLN$ 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 $S_2K = S_2L$, we directly have $\angle KMS_2 = \angle S_2ML$. 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 $\angle KRS_2 = \angle S_2RL$, then using identical angles in the cyclic quadrilaterals $RS_2MK$ and $RLS_2M$ we obtain $\angle NMS_2 = \angle KRS_2 = \angle S_2RL = \angle S_2ML$. $\blacksquare$

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 $C'$ and the line $SD$ intersects $k_2$ again in $D'$. Circles $k_1$ and $k$ meet in $C$, $C''$ and circles $k_2$ and $k$ meet in $D$, $D''$. Using the lemma we have $SC' = SC''$ and $SD' = SD''$. Let the circles $k_1$ and $k_2$ meet again in $E'$. Using a contradiction, we shall prove that $C'' = D'' = E'$.

The point $S$ lies on the chord $EE' \perp AB$ 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 $SC'' = SC' = SD' = SD''$. If $D''$ and $C''$ are different points then the triangle $\triangle SD''C''$ is isosceles and its altitude from $S$ passes through the circumcentre of $k$ and consequently (by the symmetry) the quadrilateral $CDD''C''$ is isosceles trapezoid ($SC = SD$). We can find points $A$ and $B$ as the intersection points of the axes of the segments $CC''$ and $DD''$ 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 $\angle DE'S = \angle SE'C = \alpha$ and $\angle AE'D = \beta$ then $\angle CE'D = 2\alpha - \beta$ because $2|\widehat{CD}| = |\widehat{AB}|$. Using $BD = BE'$ and $AC = AE'$ we compute the angles in the triangle $ABE'$:
$$\begin{aligned} 2\alpha + \beta &= \angle AE'C = \angle ACE' = \angle ABE' \\ 4\alpha - \beta &= \angle BE'D = \angle BDE' = \angle BAE' \end{aligned}$$
which yields to
$$180^\circ = 2\alpha + \beta + 4\alpha - \beta + 4\alpha = 10\alpha$$
and $\angle ACB = 180^\circ - \angle AE'B = 180^\circ - 4\alpha = 108^\circ$.