Question

Prove that, "The opposite angles of a cyclic quadrilateral area supplementary."

Answer 1

Construct a radius to each of the four vertices of the quadrilateral as pictured

Since the radii of the circle are all congruent, this partitions the quadrilateral into four isosceles triangles. The base angles of an isosceles triangle have the same measure. These pairs of congruent angles are labeled in the picture

The sum of the angles around the center of the circle is 360 degrees. The sum of the angles in each of the triangles is 180 degrees. So if we add up the labeled angle measures and the angles forming the circle around the center center, we get:

$2a+2b+2c+2d+360=4\left(180\right)$

$\Rightarrow a+b+c+d=180$

Alternatively, we may use the fact that the sum of angles in a quadrilateral is 360 degrees and bypass the argument with the angles making a circle.

Note that (a+b) and (c+d) are the measures of opposite angles, and we can simply group the measures in the last equation like this:

$(a+b)+(c+d)=180$

Likewise, (a+d) and (b+c) are the measures of opposite angles, and we can just rearrange the equation to see that

$(a+d)+(b+c)=180$

So, indeed, we see that the opposite angles in a cyclic quadrilateral are supplementary.