ABDC is a trapezium where AB and CD are parallel sides with AB being the smaller of the two. Sides AB, AC and BD of trap. are equal. Now a circle with centre O is drawn with AB as a chord of this while the other three sides are tangential to it. Then angle AOB is?

@scrabbler I'm unable to solve this problem even after pondering over this for a day. Can you help please.

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@scrabbler Wow, I didn't even realise that it could be done by taking it as half a regular hexagon and proceeding further from there on, learnt sth new apparently. :) What I did was, extend the lateral sides and make them meet at any point P above the AB such that APB becomes an equilateral triangle, and eventually even PCD was coming out to be an equilateral triangle. From there on it was easy by dropping a perpendicular from the point O to the base, and hence three lines meeting at the point O would each be containing the angle 120.

I had to prove the half regular hex first :) But didn't take too long :) A and B have to be points of tangency...usse hua CD as double