The figure should look like this:

We know the following about the figure:

AB = 16 so the radius of the circle equals 8

OA=OD=OE=OB because all radii of a circle are congruent.

<A and <B =60 degrees since the triangle is equilateral triangle.

Because of the isosceles triangle theorem,the measure of <ADO is equal to the measure of <DAO; and the measure of <BEO equals the meaure of <EBO.

Using substitution, the measures of the above angles equal 60 degrees.

Since the sum of the measures of the angles on one side of a line with a common vertex must equal 180 degrees, the measure of <DOE is 60 degrees.

Triangle DOE is equilateral because of the isosceles triangle theorem as well.

Therefore, the quadrilateral is made of three congruent equilateral triangles.

Using the special formulas for 30-60-90 degree triangles, the altitude of each equilateral triangle is 4*sqrt(3).

The area of one triangle is

so the area of the inscribed quadrilateral is

or square units.

Based upon the information that we gathered above, we can classify quadrilateral ABED as an isosceles trapezoid. How do we know that the segment DE is parallel to AB?

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