prove that in case of externally touching circles the radical axis is the transverse common tangent between them 
Let S1 and S2 be two circles with centers o1 and o2 respectively. By definition, radical axis of two circles is the locus of the point from which the length of the two tangents are equal. In case of externally touching circles,I read that transverse common tangent is the radical axis, but how to prove it? How can we prove that AP=AQ or BR=BS in the above diagram? I tried it by congruency but triangle o1PA and o2PA are not congruent.
Thanks in advance

It is well-known that the radical axis is a line, that, in this case, passes through the tangency point, say $T$. If $A$ is a point on the common tangent of circles $\omega_1$ and $\omega_2$, it suffices to prove that $$\text{Power}_{\omega_1}(A)=\text{Power}_{\omega_2}(A)=\iff AO_1^2-r_1^2=AO_2^2-r_2^2$$ Which is a consequence of the Pythagorean theorem in $\triangle AO_1T$ and $\triangle ATO_2$.
