Prove that AF=AB

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Question -

Draw a circle with centre O .choose a point A and cut off chord AB,BC,CD,DE,EF each equal to radius .prove that AF=AB

My try -

I know this is very simple question but I am not getting.. I tried angle chasing , try to show some congruence but none of them work...any hints ?

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If you picture a circle centred at $O$, and pick a point $A$ along the radius of the circle and cut off a chord equal to the radius at another point $B$ , and join $B$ with $O$ you will end up in an equilateral triangle of side equal to the Radius of the circle.
Carrying on this process will lead you to a regular hexagon inscribed in the circle of side equal to the radius of the circle
This would easily prove any side equal to any other leading to : $$AF= AB$$