I found the following problem.
Have a circle with centre A and another with centre B. These circles are tangent in a point named T.
P, N, B, Q are collinear points (N and Q are on second circle). P, M, A, R are collinear points (M and R are on first circle). A, T, B are collinear points. PRQ is a triangle.
Prove that if $ MN\ ||\ RQ \Rightarrow \triangle PAB$ isosceles.
I also have the following figure:
Hope my english translation is intelligible!
I redrew the figure to have $ MN\ ||\ RQ $, but I have no idea how to solve that problem. I tried to find some relations with the MRQN trapezium and AB as its middle line but with no success.
How can this be solved?
Thank you!
The theorem is false (see counterexample below). Are you sure the circles can have different radii?
EDIT, for those who don't believe their eyes.
Draw the two circles at will, tangent at $T$. Draw any diameter $MR$ in the circle of center $A$, then from $M$ and $R$ draw two lines parallel to $AB$. The intersections of these lines with the circle of center $B$ are the endpoints of two diameters: let $NQ$ be one of them. If lines $MR$ and $NQ$ meet at $P$, then triangles $PMN$, $PAB$, $PRQ$ are similar and $PA:PB=MA:NB$. Hence $PA=PB$ only if the circles have the same radii.
On the other hand, $AB=AP$ implies $MN=MP=TB$, and $MNBT$ is then a rhombus. We can then have $AB=AP$ only if $BT=MT$.