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

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

#### Proving Supplementary Angles

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

First note that mDEBˆ+mDCBˆ=360\begin{align*}m \widehat{DEB} + m \widehat{DCB}=360^\circ\end{align*} because these two arcs make a full circle. 2mDEB=mDEBˆ\begin{align*}2m \angle DEB = m \widehat{DEB}\end{align*} and 2mDCB=mDCBˆ\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, 2mDEB+2mDCB=360\begin{align*}2m \angle DEB + 2m \angle DCB = 360^\circ\end{align*}. Divide by 2 and you have mDEB+mDCB=180\begin{align*}m \angle DEB + m \angle DCB = 180^\circ\end{align*}. Therefore, DEB\begin{align*}\angle DEB\end{align*} and DCB\begin{align*}\angle DCB\end{align*} are supplementary.

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

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.

#### Finding a Point of Intersection

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

CDE\begin{align*}\angle CDE \end{align*} is an exterior angle of ΔFCD\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 mCDE=mFCD+mF\begin{align*}m \angle CDE = m \angle FCD + m \angle F\end{align*}. Quadrilateral BCDE\begin{align*}BCDE\end{align*} is cyclic, so CDE\begin{align*}\angle CDE\end{align*} and B\begin{align*}\angle B\end{align*} must be supplementary. This means that CDE\begin{align*}\angle CDE\end{align*} and F\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.

#### Solving for Unknown Values

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

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

### Examples

#### Example 1

Earlier, you were given a problem about a rhombus.

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

#### Example 2

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

BCD\begin{align*}\angle BCD\end{align*} is the inscribed angle of DEBˆ\begin{align*}\widehat{DEB}\end{align*}. This means that the measure of the arc is twice the measure of the angle. mDEBˆ=872=174\begin{align*}m \widehat{DEB}=87^\circ \cdot 2 = 174^\circ\end{align*}. Since mBEˆ=76\begin{align*}m \widehat{BE}=76^\circ\end{align*}, mDEˆ=17476=98\begin{align*}m \widehat{DE}=174^\circ - 76^\circ = 98^\circ\end{align*}.

#### Example 3

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

BCD\begin{align*}\angle BCD\end{align*} and DEB\begin{align*}\angle DEB\end{align*} are opposite angles of a cyclic quadrilateral so they are supplementary. mDEB=18087=93\begin{align*}m \angle DEB = 180^\circ - 87^\circ = 93^\circ\end{align*}.

#### Example 4

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

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

### Review

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.

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### Vocabulary Language: English

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.