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Inscribed Quadrilaterals in Circles

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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:

https://www.youtube.com/watch?v=vk0Td5MgW90 Brightstorm: 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.

Note: You will learn more about proof by contradiction in future courses!

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} .

Answers:

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.

Image Attributions

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