Drawing the picture, one gets more insight about the
position of the lines $PQ$, $RS$, $XY$, this insight solves the problem.
Here is the pic:
The statement of the problem is so that points are not introduced in the "right order". But let us resume, to see what we have.
First statement:
The following data is given. There are three circles
$\omega_1,\omega_2,\omega_3$ with centers in $O_1,O_2,O_3$ respectively.
- The circles $\omega_1, \omega_2$ intersect in $X,Y$.
- The circles $\omega_2, \omega_3$ intersect in $P,Q$.
- The circles $\omega_1, \omega_3$ intersect in $R,S$.
As a coincidence,
the point $O_1$ is given to be on the line $PQ=:l_1$,
and $O_2$ is given to be on the line $RS=:l_2$.
We have to show that $O_3$ is on the third line $XY=:l_3$.
Proof: See the restatement below.
$\square$
We restate equivalently, so that the given coincidence is a part of a construction, which is a consequence of the given data. So here is an equivalent way to restate. We observe that $O_1O_3$ is the side bisector of the segment $RS$, so in particular it is perpendicular on $RS=l_2$.
Similarly, $O_2O_3$ is perpendicular on $PQ=l_1$. We use this information first.
Second (equivalent) statement:
The following data is given. There is a triangle $\Delta O_1O_2O_3$,
and let $H$ be its orthocenter. We draw the heights (as lines)
$l_1,l_2,l_3$ through $H$.
Let $\omega_3$ be a circle centered in $O_3$, which (has a sufficiently big radius so that it) intersects $l_1$ in $P,Q$, and $l_2$ in $R,S$.
- Then $O_1$ is on the side bisector of $RS$, so there is a circle $\omega_1$ centered in $O_1$ passing through $R,S$.
- And similarly $O_2$ is on the side bisector of $PQ$, so there is a circle $\omega_2$ centered in $O_2$ passing through $P,Q$.
We furthermore assume that $\omega_1,\omega_2$ intersect, let $X,Y$ be the points of intersection.
Then the line $XY$ coincides with the line (height) $l_3$.
Proof: (The same figure above applies.)
See the restatement below.
$\square$
We remove objects as much as possible from the picture. We need only the
triangle $\Delta O_1O_2O_3$ and its heights. Having the height through
$O_1$, we do not need both points $P,Q$, one of them is enough.
So we keep only one point in each pair.
We no longer draw the two circles through $P$, it is enough to know the property relating $P$ and $R$, namely $O_3P=O_3R$. So we restate:
Third (equivalent) statement:
In $\Delta O_1O_2O_3$, we project the vertices on opposite sides into
$K_1,K_2,K_3$, and let $H=O_1K_1\cap O_2K_2\cap O_3K_3$ be orthocenter.
We pick points $P$ on $O_1K_1$, $R$ on $O_2K_2$, and $T$ on $O_3K_3$,
so that two of the three relations are holding:
$$
\begin{aligned}
O_3R &= O_3P\ ,\\
O_2P &= O_2T\ ,\\
O_1T &= O_1R\ .
\end{aligned}
$$
Then the third relations is also true.
Proof of the equivalence of the second and third statements:
In the situation from the second statement,
keep $\Delta O_1O_2O_3$, and the points $P,R$ on the heights from $O_1,O_2$, with same distance to $O_3$, so the first relation above is satisfied.
Construct $T$ on the third height satisfying a second relation among the above three. Then the third is also satisfied. So $T$ is in the intersection $\omega_1\cap\omega_2$, being thus either the point $X$ or the point $Y$ from the second statement. So $X$ or $Y$ is on the $O_3$-height, so from $XY\perp O_1O_2$ the other point is on it, too.
$\square$
Proof of the third statement:
We have $O_3P^2-O_3H^2=(O_3K_1^2+PK_1^2)-(O_3K_1^2+HK_1^2)=PK_1^2-HK_1^2$.
Writing similar relations also for $R,T$, we obtain:
$$
\begin{aligned}
O_3R^2 - O_3P^2 &= (RK_2^2-HK_2^2) - (PK_1^2-HK_1^2)\ ,\\
O_2P^2 - O_2T^2 &= (PK_1^2-HK_1^2) - (TK_3^2-HK_3^2)\ ,\\
O_1T^2 - O_1P^2 &= (TK_3^2-HK_3^2) - (RK_2^2-HK_2^2)\ .
\end{aligned}
$$
Adding the relations, we obtain zero on the R.H.S. - so if two of the three relations from the statement are true, the third one is also true.
