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## Quadrilaterals with every vertex on a circle and opposite angles that are supplementary.

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One angle of a rhombus is \begin{align*}30^\circ\end{align*}. Can this rhombus be inscribed in a circle?

#### Watch This

Watch the first part of this video that discuses cyclic quadrilaterals:

#### Guidance

A quadrilateral is said to be inscribed in a circle if all four vertices of the quadrilateral lie on the circle. Quadrilaterals that can be inscribed in circles are known as cyclic quadrilaterals. The quadrilateral below is a cyclic quadrilateral.

Not all quadrilaterals can be inscribed in circles and so not all quadrilaterals are cyclic quadrilaterals. A quadrilateral is cyclic if and only if its opposite angles are supplementary. You will prove this theorem in Examples A and B.

Example A

Consider the cyclic quadrilateral below. Prove that \begin{align*}\angle DEB \end{align*} and \begin{align*}\angle DCB\end{align*} are supplementary.

Solution: First note that \begin{align*}m \widehat{DEB} + m \widehat{DCB}=360^\circ\end{align*} because these two arcs make a full circle. \begin{align*}2m \angle DEB = m \widehat{DEB}\end{align*} and \begin{align*}2m \angle DCB = m \widehat{DCB}\end{align*} because the measure of an inscribed angle is half the measure of its intercepted arc. By substitution, \begin{align*}2m \angle DEB + 2m \angle DCB = 360^\circ\end{align*}. Divide by 2 and you have \begin{align*}m \angle DEB + m \angle DCB = 180^\circ\end{align*}. Therefore, \begin{align*}\angle DEB\end{align*} and \begin{align*}\angle DCB\end{align*} are supplementary.

Example B

Consider the quadrilateral below. Assume that \begin{align*}\angle B\end{align*} and \begin{align*}\angle F\end{align*} are supplementary, but that \begin{align*}F\end{align*} does NOT lie on the circle. Find a contradiction. What does this prove?

Solution: One method of proof is called a proof by contradiction. With a proof by contradiction you prove that something cannot not be true. Therefore, it must be true. Here, you are attempting to prove that it is impossible for a quadrilateral with opposite angles supplementary to not be cyclic. Therefore, such a quadrilateral must be cyclic.

Assume that \begin{align*}\angle B\end{align*} and \begin{align*}\angle F\end{align*} are supplementary, but \begin{align*}F\end{align*} is not on the circle. Find the point of intersection of \begin{align*}\overline{EF}\end{align*} and the circle and call it \begin{align*}D\end{align*}. Connect \begin{align*}D\end{align*} with \begin{align*}C\end{align*}.

\begin{align*}\angle CDE \end{align*} is an exterior angle of \begin{align*}\Delta FCD\end{align*}, so its measure is equal to the sum of the measures of the remote interior angles of the triangle. This means that \begin{align*}m \angle CDE = m \angle FCD + m \angle F\end{align*}. Quadrilateral \begin{align*}BCDE\end{align*} is cyclic, so \begin{align*}\angle CDE\end{align*} and \begin{align*}\angle B\end{align*} must be supplementary. This means that \begin{align*}\angle CDE\end{align*} and \begin{align*}\angle F\end{align*} must be congruent because they are both supplementary to the same angle.

The two highlighted statements are a contradiction -- they cannot both be true. This means that your original assumption cannot exist. You cannot have a quadrilateral with opposite angles supplementary that is not cyclic. So, if opposite angles of a quadrilateral are supplementary then the quadrilateral must be cyclic.

Example C

Solve for \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Solution: Opposite angles are supplementary, so \begin{align*}90^\circ + x^\circ = 180^\circ\end{align*} and \begin{align*}100^\circ + y^\circ = 180^\circ\end{align*}. This means \begin{align*}x=90\end{align*} and \begin{align*}y=80\end{align*}.

Concept Problem Revisited

One angle of a rhombus is \begin{align*}30^\circ\end{align*}. Can this rhombus be inscribed in a circle?

Opposite angles of a rhombus are congruent. If a rhombus has a \begin{align*}30^\circ\end{align*} angle then it has one pair of opposite angles that are each \begin{align*}30^\circ\end{align*} and one pair of opposite angles that are each \begin{align*}150^\circ\end{align*}. Opposite angles are not supplementary so this rhombus cannot be inscribed in a circle.

#### Vocabulary

A quadrilateral is said to be inscribed in a circle if all four vertices of the quadrilateral lie on the circle. Quadrilaterals that can be inscribed in circles are known as cyclic quadrilaterals.

A proof by contradiction assumes the opposite of what you are trying to prove and shows that a contradiction will exist with such an assumption. Therefore, what you are trying to prove must be true.

#### Guided Practice

1. Find \begin{align*}m \widehat{DE}\end{align*}.

2. Find \begin{align*}m \angle DEB\end{align*}.

3. Find \begin{align*}m \widehat{CB}\end{align*}.

1. \begin{align*}\angle BCD\end{align*} is the inscribed angle of \begin{align*}\widehat{DEB}\end{align*}. This means that the measure of the arc is twice the measure of the angle. \begin{align*}m \widehat{DEB}=87^\circ \cdot 2 = 174^\circ\end{align*}. Since \begin{align*}m \widehat{BE}=76^\circ\end{align*}, \begin{align*}m \widehat{DE}=174^\circ - 76^\circ = 98^\circ\end{align*}.

2. \begin{align*}\angle BCD\end{align*} and \begin{align*}\angle DEB\end{align*} are opposite angles of a cyclic quadrilateral so they are supplementary. \begin{align*}m \angle DEB = 180^\circ - 87^\circ = 93^\circ\end{align*}.

3. A full circle is \begin{align*}360^\circ\end{align*}. \begin{align*}m \widehat{CB}=360^\circ - 60^\circ - 98^\circ - 76^\circ = 126^\circ\end{align*}.

#### Practice

1. What is a cyclic quadrilateral?

2. A quadrilateral is cyclic if and only if its opposite angles are __________________.

3. Find \begin{align*}m \angle B\end{align*}.

4. Find \begin{align*}m \angle E\end{align*}.

5. Find \begin{align*}m \angle D\end{align*}.

6. Find \begin{align*}m \widehat{CD}\end{align*}.

7. Find \begin{align*}m \widehat{DE}\end{align*}.

8. Find \begin{align*}m \angle CBE\end{align*}.

9. Find \begin{align*}m \angle CEB\end{align*}.

10. Solve for \begin{align*}x\end{align*}.

11. Solve for \begin{align*}y\end{align*}.

12. Solve for \begin{align*}x\end{align*}.

13. Solve for \begin{align*}y\end{align*}.

14. If a cyclic quadrilateral has a \begin{align*}90^\circ\end{align*} angle, must it be a square? If yes, explain. If no, give a counter example.

15. Use the picture below to prove that angles \begin{align*}B\end{align*} and \begin{align*}D\end{align*} must be supplementary.

### Vocabulary Language: English

Inscribed Polygon

Inscribed Polygon

An inscribed polygon is a polygon with every vertex on a given circle.

The Inscribed Quadrilateral Theorem states that a quadrilateral can be inscribed in a circle if and only if the opposite angles of the quadrilateral are supplementary.

A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle.