Geometry question on trapezium A trapezium $ABCD$, in which $AB$ is parallel to $CD$, is inscribed in a circle with center $O$. Suppose the diagonals $AC$ and $BD$ of the trapezium intersect at $M$, and $OM=2$. How do I find the difference between parallel sides : * if angle $A\hat{M}B = 60$° * if angle $A\hat{M}D = 60$°

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It should be clear that $\triangle MAB$ and $\triangle MCD$ are equilateral.

After dropping the perpendicular from O to AC at P, (as hinted by @Rigel), $\triangle OPM$ is a 30-60-90 special triangle with the lengths as shown.

Let $PC = t$. Then, $DC = MC = t + \sqrt 3$

$AB = AM = PA – PM = t - \sqrt 3$

The said difference $= …. = 2 \cdot \sqrt 3$, after the elimination of t.

![enter image description here](

Referring to the second diagram, we can apply the same logic to the second case with the following exceptions:-

1. The special angled triangle is still a $30-60-90# one but with different orientation.

2. $\triangle \; s \; MAB$ and $MCD$ are now isosceles with equal vertical angle $= 120^0$ and the legs are s – 1 and s + 1 respectively.




We can apply the cosine law to both triangles to find AB and also CD. Then calculate CD – AB.