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A circle with center O is inscribed in a quadrilateral ABCD. AB// CD, $\angle BCD$ = 60 and $\angle ADC$ = 40. What is the measure of $\angle$ AOC ?

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My Try: If E, F, G, H are the points on AB, BC, CD, DA where those lines touch the circle, then OE is perpendicular to AB, OF perp to BC etc.

Also, when two tangents meet at a point outside the circle, the lengths of the two tangents are equal: AE = AH, BE=BF etc.

So if I draw these in, you get 4 pairs of congruent right-angled triangles OAE, OAH etc. Since they're congruent, the angles ADC and BCD are bisected by OD, OC.

I have no idea how to proceed from here. Could someone do the remaining angles part? Please

Answer of this problem: 140.

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Since AB and CD are parallel and you are given 2 other angles, you should be able to determine the remaining 2 angles in the quadrilateral. You might also assume that the center of the inscribed circle fits on the bisectors of the two given angles. With simple geometric logic, you should be able to figure out the rest.

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