Let ABCD be a convex quadrilateral for which the circle of diameter AB is tangent to the line CD. Show that the circle of diameter CD is tangent to the line AB if and only if the lines BC and AD are parallel.
My solution:
If the $AD = BC$ the result is obvious. Suppose WLOG $BC < AD$ and extend $CD$ to meet $AB$ at $P$.
Let $CP=h$, $BP=m$, $CD/2=r$, and $AB/2=R$.
Let the centre of the circle with diameter $AB$ be $S$ and the foot of the perpendicular from $S$ to $CD$ be $T$.
Let the centre of the circle with diameter $CD$ be $Q$ and the foot of the perpendicular from $Q$ to $AB$ be $N$. Let $QN=x$.
The lines $BC$ and $AD$ are parallel iff PBC and PAD are similar, which is the case iff
$\frac{h}{m}=\frac{h+2r}{m+2R}\iff\frac{h}{m}=\frac{r}{R}$
Now PQN and PST are similar:
$\therefore \frac{h+r}{x}=\frac{m+R}{R}\iff x=R\frac{h+r}{m+R}$
from which it follows that $x=r$ iff PBC and PAD are similar.
QED
I'm wondering if this solution would get full credit, and welcome any suggestions for improvement. Thanks.