# Quadrilaterals Inscribed in a Circle: Opposite Angles Theorem

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• 1:00 Opposite Angles Theorem
• 1:35 Example Problem
• 3:45 Lesson Summary
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Lesson Transcript
Instructor: Elizabeth Foster

Elizabeth has been involved with tutoring since high school and has a B.A. in Classics.

Have you ever looked at your geometry book and thought, 'Hey, you know what these pictures need? More shapes!' No? Well, even if you never thought it, that's what you're going to get in this lesson: double-shape action with circles and quadrilaterals.

A quadrilateral is any shape with four sides and four angles. Quadrilaterals include squares, rectangles, trapezoids, and just random shapes that happen to have four sides and four angles.

Now imagine trying to draw a circle around each of these shapes. See how some of them touch the circle with all four corners, while other ones have one or two corners that are just kind of floating around? If you can draw a circle around the quadrilateral to touch all four corners of the shape to the outside of the circle, it's called inscribing the quadrilateral in the circle.

A cyclic quadrilateral is a quadrilateral inscribed in a circle. Not all quadrilaterals are cyclic. But quite a few of them are.

## Opposite Angles Theorem

But why would anyone want to draw a circle around a quadrilateral? Why does it matter whether the quadrilateral is cyclic or not? Because if you can inscribe it in a circle, you know something about the quadrilateral. In a cyclic quadrilateral, opposite angles are supplementary.

If a pair of angles are supplementary, that means they add up to 180 degrees. So if you have any quadrilateral inscribed in a circle, you can use that to help you figure out the angle measures.

## Example Problem

If D = 3E and triangle AFB is equilateral, what is the measure of angle C?

We can see that this is basically a quadrilateral with a line through the middle dividing it into two triangles. Since it's a cyclic quadrilateral, the opposite angles must be supplementary. In this drawing, we have two pairs of opposite angles:

• A + D = 180
• (F + E) + (B + C) = 180

We also know that if the triangle AFB is equilateral, then all the angles are equal to 60 degrees. By plugging these values into the supplementary angles equation, we can say that 120 + C + E = 180.

We also know that C + E + D = 180 because all the angles in a triangle add up to 180. The problem tells us that D = 3E. If we substitute 3E for D in the triangle equation, we get C + E + 3E = 180, or C + 4E = 180.

Rearrange that just a little, so it reads C = 180 - 4E. Now we can plug in for C in our original equation: 120 + (180 - 4E) + E = 180. We'll simplify: 120 - 3E = 0. 3E = 120. E = 40.

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