Two circles tangent in a point. Having two parallel segments on the circles find an isoscele triangle. 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: ![An example of a 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. ![My draw.]( How can this be solved? Thank you!

The theorem is false (see counterexample below). Are you sure the circles can have different radii?

![enter image description here](

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 $BT=MT$.