Let $\omega_1$ and $\omega_2$ be two orthogonal circles (definition given below). Let the centre of $\omega_1$ be $O$ . Diameter $AB$ of $\omega_1$ is selected so that $B$ lies strictly inside $\omega_2$. The two circles tangent to $\omega_2$ through both $O$ and $A$ touch $\omega_2$ at $F$ and $G$ . Prove that quadrilateral $FOGB$ is cyclic.
NOTE:
Let two circles with centres $P$ and $Q$ intersect at $X$ and $Y$. Then the two circle are orthogonal iff $\angle PXQ = \angle PYQ = 90 \circ $ .
Centre of circle, an antipode and tangency points are cyclic
Re: Centre of circle, an antipode and tangency points are cy
Hint:
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