D is an arbitrary point on ⊙O1 ∠OCQ=∠OCP using sin law we obtain sin∠CBO=sin∠CPO So ∠CBO=∠CPO or ∠CBO+∠CPO=180 But ∠CDO+∠CPO=180 ∠CBO >∠CDO =〉∠CBO=∠CPO =〉CB=CP =〉AB=PQ

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Draw $ D$ such that $ OD$ is a diameter of the first circle. Then $ \angle DPO$ is right so $ DP$ is tangent to the second circle. Similarly, $ DQ$ is tangent to the second circle. We have \begin{align*} \angle AOP &= \angle DPC \\ &= \angle DQC \\ &= \angle QOB. \end{align*} This means $ \angle AOB=\angle POQ$ or $ AB=PQ$ as desired. [geogebra]31aa5346dea7f3a0b753f9a66c361e20d6e7840b[/geogebra]

From $PCQO$ cyclic we get $\widehat{OPA}=\widehat{BQO}$ $(1)$, triangles $AOP$ and $BOQ$ are isosceles and, with $AO=BO$ they are congruent, hence $AB=PQ$, done. Best regards, sunken rock