Two circles touches each other externally at the point O. .. I am stuck on the following elementary problem that says:

Two circles touches each other externally at the point O. If PQ and RS are two diameters of these two circles respectively and $PQ || RS$ (Where || indicates parallel),then prove that P,O,S  are collinear. 


My Try: 
 I added the points P,O,S and also the points Q,O; ans R,O. 
From the given condition , $\angle POQ=90^{\circ}=\angle ROS$. [since,Any angle inscribed in a semi-circle is a right angle.]
Now,from the figure we see  $\angle POR=180^{\circ}-90^{\circ}=\angle QOS$.
And hence,$\angle POQ+\angle QOS=90^{\circ}+90^{\circ}=180^{\circ}=\angle POS$.
Hence ,we can conclude P,O,S are collinear.
Can someone verify it ? Am I right? 
Thanks in advance for your time. 

$ POQ,ROS$ are right angles because they are contained in a semi-circle, by Thales theorem.
EDIT1:
My bad, an error occured before. 
Imagine two disjunct  right triangles hinged at right angle vertex with discontinuous slope. Unless alternate angles at diameters are equal a same line continuity  cannot be established. When cyan right triangle at right is rotated so that $ \beta = B $ these are alternate angles. So the transversal cutting line through $O$ has to be single continuous cutting line.

