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I would be very grateful if you can help me with this problem. I've constructed the median ON, N ∈ BC, and I was able to find that the triangle OCN is isosceles (height and median coincide). Probably I have to prove it is equilateral, and the only way I can think of it is to show that there is an angle 60 degrees. I can't find a way to do it. Thanks in advance!

ABCD (AB || CD) is a rectangular trapeze, AD ⊥ AB. A circle with center midpoint O of AD is in contact with BC at M (touching). If AB = 36 cm CD = 12 cm, find the length of OC.

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Since $$\measuredangle DCB+\measuredangle ABC=180^{\circ},$$ we obtain: $$\measuredangle COB=180^{\circ}-\frac{1}{2}\left(\measuredangle DCB+\measuredangle ABC\right)=90^{\circ},$$ which gives $$OM^2=CM\cdot MB=DC\cdot AB=12\cdot36.$$ Id est, by the Pythagoras's theorem $$OC=\sqrt{OC^2}=\sqrt{OD^2+DC^2}=\sqrt{OM^2+DC^2}=\sqrt{12\cdot36+12^2}=24.$$

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