, 2014

Claim: The point $E$ is the intersection point of the line parallel to $AD$ through $B$ and the line $CD$.

Let $E$ be the intersection point of the line parallel to $AD$ through $B$ and the line $CD$. Firstly, we will show that $E$ is on the segment $CD$. Since $AB \le AD$ and $BC \le CD$ (because $AB$ and $CD$ are the shortest and longest sides, respectively), then $\angle ABD \ge \angle ADB$ and $\angle CBD \ge \angle BDC$. Therefore $\angle ABC = \angle ABD + \angle DBC \ge \angle ADB + \angle BDC = \angle ADC$. Similarly, we will have that $\angle BAD \ge \angle BCD$. Thus, $\angle ABC + \angle BAD \ge \angle BCD + \angle CDA$, and then $\angle ABC + \angle BAD \ge 180^\circ$. Moreover, one can show that $\angle ABC + \angle BAD > 180^\circ$. (If $\angle ABC + \angle BAD = 180^\circ = \angle BCD + \angle CDA$, we would have that $\angle ABC = \angle ADC$ and $\angle BAD = \angle BCD$, and then $\square ABCD$ would be a parallelogram; contradicting with the fact that $AB < CD$.) Therefore, we can move the point $C$ along the line $CD$ closer to $D$ to the point $E'$, making the angle $ABC$ smaller, so that $\angle E'BA + \angle BAD = 180^\circ$, i.e., $BE' \parallel AD$. Thus, $E'$ is the same as the point $E$ stated above, and lies on the segment $CD$.

Secondly, we will show that for an arbitrary point $P$ different from $E$ on the segment $CD$, $\angle O_1PO_2 = \angle APB$.

Case 1: $\angle ADP < 90^\circ$. Thus $\angle BEC < 90^\circ$, and have that $O_1$ and $C$ are on the opposite sides of $PA$, and $O_2$ and $C$ are on the opposite sides of $PB$. By simple angle chasing (using angles at centres) we get that
$$\angle APO_1 = 90^\circ - \angle ADP = 90^\circ - \angle BEC = \angle BPO_2.$$
Therefore $\angle O_1PO_2 = \angle APB$.

Case 2: $\angle ADP \ge 90^\circ$. Similarly, we get $\angle BEC \ge 90^\circ$ but now $O_1$ and $C$ are on the same side of $AP$, and $O_2$ and $C$ are on the same side of $BP$. So by angle chasing we get
$$\angle APO_1 = \angle ADP - 90^\circ = \angle BEC - 90^\circ = \angle BPO_2.$$
Therefore $\angle O_1PO_2 = \angle APB$.

Finally, using Law of Sine for $AP$ and $BP$ on circles $O_1$ and $O_2$, respectively, we get
$$\frac{AP}{BP} = \frac{2O_1P \sin(\angle ADP)}{2O_2P \sin(\angle BEP)} = \frac{O_1P}{O_2P}.$$
Thus we have that $\triangle APB \sim \triangle O_1PO_2$, and to conclude that
$$O_1O_2 = \frac{AB}{AP} \cdot O_1P = \frac{AB}{2\sin(\angle ADC)}$$
which is a constant independent of a point $P$. $\square$