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Tangents $OA$ and $OB$ are drawn to a circle from an external point $O$. Through the point $A$, a chord $AC$ is drawn parallel to the tangent $OB$ and $OC$ passes through the circle at $E$.

I am required to show that A$F$ bisects $OB$.

My approach thus far has been to observe that triangles $AFO$ and $OFE$ are similar, and also $AEB$ is similar to $BEO$.

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However I am not able to reach an appropriate proof... Please help!

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  • $\begingroup$ I am required to show hat AF bisects OB. So yes, equivalently F would be the midpoint of BO. $\endgroup$ – Hugh Entwistle Jul 22 '18 at 9:48
  • $\begingroup$ Math.SE demands you to show at least some work on your question. What have you tried so far? $\endgroup$ – Cargobob Jul 22 '18 at 9:50
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    $\begingroup$ I detailed above, that I have been able to prove that Triangles AFO and OFE are similar. And that triangles ABE and EBO are similar. Working out was omitted, but I can give you the working if you so desire. $\endgroup$ – Hugh Entwistle Jul 22 '18 at 9:52
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Since $\angle EAO = \angle EOF$ we see that the line $FO$ is tangent to circumcircle $(AEO)$, so by PoP with respect to $F$ and $(AEO)$ we have $$FO^2 = FE\cdot FA$$

and by PoP on a given circle we have $$FB^2 = FE\cdot FA$$

so $FB = FO$.

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  • $\begingroup$ Is PoP for "Power of Point" ? (sorry, not accustomed to this abbreviation). $\endgroup$ – Jean Marie Jul 22 '18 at 18:01
  • $\begingroup$ Yes, a saw it elsewhere $\endgroup$ – Aqua Jul 22 '18 at 18:02

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