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One angle of a rhombus is $30^\circ$ . 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  $\angle DEB$ and  $\angle DCB$ are supplementary.

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

Example B

Consider the quadrilateral below. Assume that  $\angle B$ and  $\angle F$ are supplementary, but that  $F$ 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  $\angle B$ and  $\angle F$ are supplementary, but  $F$ is not on the circle. Find the point of intersection of  $\overline{EF}$ and the circle and call it $D$ . Connect  $D$ with $C$ .

$\angle CDE$  is an exterior angle of $\Delta FCD$ , so its measure is equal to the sum of the measures of the remote interior angles of the triangle. This means that $m \angle CDE = m \angle FCD + m \angle F$ . Quadrilateral  $BCDE$ is cyclic, so  $\angle CDE$ and  $\angle B$ must be supplementary. This means that  $\angle CDE$ and  $\angle F$ 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  $x$ and $y$ .

Solution: Opposite angles are supplementary, so  $90^\circ + x^\circ = 180^\circ$ and $100^\circ + y^\circ = 180^\circ$ . This means $x=90$ and $y=80$ .

Concept Problem Revisited

One angle of a rhombus is $30^\circ$ . Can this rhombus be inscribed in a circle?

Opposite angles of a rhombus are congruent. If a rhombus has a  $30^\circ$ angle then it has one pair of opposite angles that are each  $30^\circ$ and one pair of opposite angles that are each $150^\circ$ . 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 $m \widehat{DE}$ .

2. Find $m \angle DEB$ .

3. Find $m \widehat{CB}$ .

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

2.  $\angle BCD$ and  $\angle DEB$ are opposite angles of a cyclic quadrilateral so they are supplementary. $m \angle DEB = 180^\circ - 87^\circ = 93^\circ$ .

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

Practice

1. What is a cyclic quadrilateral?

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

3. Find $m \angle B$ .

4. Find $m \angle E$ .

5. Find $m \angle D$ .

6. Find $m \widehat{CD}$ .

7. Find $m \widehat{DE}$ .

8. Find $m \angle CBE$ .

9. Find $m \angle CEB$ .

10. Solve for $x$ .

11. Solve for $y$ .

12. Solve for $x$ .

13. Solve for $y$ .

14. If a cyclic quadrilateral has a  $90^\circ$ angle, must it be a square? If yes, explain. If no, give a counter example.

15. Use the picture below to prove that angles  $B$ and  $D$ must be supplementary.