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Polygons with a different number of sides can be inscribed in a circle. In this lesson, inscribed quadrilaterals, or polygons with four sides, will be explored. Furthermore, three main properties of inscribed quadrilaterals will be investigated. ### Catch-Up and Review

Write the answer without the degree symbol.

**Here are a few recommended readings before getting started with this lesson.**

Try your knowledge on these topics.

a Pair each geometric object with its definition.

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b In the circle, ∠MON measures $70_{∘}.$ Find the measure of the corresponding inscribed angle.

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c Calculate the sum of the arc measures.

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d On the circle, the measures of all arcs except one are given. Find the measure of that arc and the measure of the inscribed angle ∠A.

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e Use the Polygon Interior Angles Theorem to calculate the missing angle measure of DEFG.

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An inscribed quadrilateral is a quadrilateral whose vertices all lie on a circle. It can also be called a *cyclic quadrilateral*.
### Proof

### Part 1

The sum of the measures of ∠A and ∠C is $180_{∘}.$ This means that these angles are supplementary. By the same logic, it can be also proven that ∠B and ∠D are supplementary. This concludes the proof of Part 1. ### Part 2

It was assumed that ABCD has supplementary opposite angles. Therefore, ∠D and ∠B are supplementary angles.
By the Transitive Property of Equality, the above equations imply that ∠AEC and ∠D have equal measures.
However, this is not possible. The reason is that the measure of the exterior angle ∠AEC of △EDC can not be the same as the measure of the interior angle ∠D.

In the diagram above, all four vertices of ABCD lie on a circle. Therefore, ABCD is a cyclic quadrilateral.

A quadrilateral can be inscribed in a circle if and only if its opposite angles are supplementary.

Based on the diagram above, the following relations hold true.

This theorem will be proven in two parts.

- If a quadrilateral can be inscribed in a circle, then its opposite angles are supplementary.
- If the opposite angles of a quadrilateral are supplementary, then it can be inscribed in a circle.

Consider a circle and an inscribed quadrilateral ABCD.

Notice that $BCD$ and $BAD$ together span the entire circle. Therefore, by the Arc Addition Postulate, the sum of their measures is $360_{∘}.$$mBCD+mBAD=360_{∘} $

From the diagram, it can be seen that $BCD$ and $BAD$ are intercepted by ∠A and ∠C, respectively.
By the Inscribed Angle Theorem, the measure of each of these inscribed angles is half the measure of its intercepted arc.
$m∠A=21 mBCDm∠C=21 mBAD $

The above equations can be simplified by multiplying both sides of each equation by 2.
$2m∠A=mBCD2m∠C=mBAD $

Next, 2m∠A and 2m∠C can be substituted for $mBCD$ and $mBAD,$ respectively, into the equation found earlier. $mBCD+mBAD=360_{∘}$

SubstituteII

$mBCD=2m∠A$, $mBAD=2m∠C$

$2m∠A+2m∠C=360_{∘}$

FactorOut

Factor out 2

$2(m∠A+m∠C)=360_{∘}$

DivEqn

$LHS/2=RHS/2$

$m∠A+m∠C=180_{∘}$

This part of the proof will be proven by contradiction. Suppose that ABCD is a quadrilateral that has supplementary opposite angles, but ABCD is **not** cyclic.

Since ABCD is not cyclic, the circle that passes through A, B, and C, does **not** pass through D. Let E be the point of intersection of AD and the circle. Consider the quadrilateral ABCE.

This contradiction proves that the initial assumption was **false**, and ABCD is a cyclic quadrilateral. Note that a similar argument can be used if D lies *inside* the circle. The proof of Part 2 is now complete.

The Inscribed Quadrilateral Theorem can be used to identify whether a quadrilateral is cyclic.

Tiffaniqua is given a quadrilateral JKLM. She wants to draw a circle that passes through all the vertices, but she does not know if it is possible. For that reason, she decided to measure the angles of JKLM.

Help Tiffaniqua determine whether it is possible to inscribe JKLM into a circle.{"type":"choice","form":{"alts":["Yes","No"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":1}

Compare the sums of the opposite angles' measures.

The Inscribed Quadrilateral Theorem can be used to determine whether JKLM is cyclic. According this theorem, the opposite angles of the quadrilateral need to be supplementary. Calculate the sum of opposite angles' measures and see if it is true.

Pair 1 | Pair 2 | |
---|---|---|

Opposite Angles | ∠J and ∠L | ∠K and ∠M |

Sum | $99_{∘}+74_{∘}=173_{∘}×$ | $105_{∘}+82_{∘}=187_{∘}×$ |

The sum of the angle measures in each pair is not equal to $180_{∘}.$ Therefore, neither ∠J and ∠L nor ∠K and ∠M are supplementary. This finding implies that JKLM is **not** a cyclic quadrilateral.

Find the measure of ∠D. Write your answer without the degree symbol.

On the diagram below, one side of a cyclic quadrilateral ABCD is extended to E. As a result, ∠ADE — the exterior angle of ABCD — is formed.

In this case, ∠ABC is said to be the opposite interior angle. The relationship between these angles is described by the *Cyclic Quadrilateral Exterior Angle Theorem*.

If a side of a cyclic quadrilateral is extended, then the exterior angle is congruent to the opposite interior angle.

Based on the diagram above, the following relation holds true.

∠ABC≅∠ADE

Consider an inscribed quadrilateral with one side extended to point E.

From the diagram, it can be observed that ∠ADE and ADC form a linear pair. Therefore, these angles are supplementary, which means that the sum of their measures is $180_{∘}.$∠ABC≅∠ADE

This relation is illustrated on the diagram below.

By the same logic, this theorem can be proven for any other extended side of ABCD. The proof is now complete.

Davontay wants to go to a concert, but his parents say that he has to finish his homework first. In the last math exercise, he is asked to find the values of all variables.

Help Davontay solve the last exercise so that he can go to the concert.

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Identify the exterior angles to the inscribed quadrilateral WXYZ and the opposite interior angles. Then use the property that states these angles are congruent.

By observing the diagram, ∠VXY and ∠YXW can be recognized to form a linear pair, so they are supplementary angles.
By substituting $84_{∘}$ for m∠VXY and $a_{∘}$ for m∠YXW into the equation, the value of a can be calculated.
The value of a is 96. Next, notice that ∠VXY is the exterior angle of WXYZ, while ∠WZY is the opposite interior angle.
Therefore, these angles are congruent.
The measure of VXY is $84_{∘}$ and the measure of ∠WZY is $3b_{∘}.$ By substituting these values and solving the equation, the value of b can be found.
Therefore, the value of b is 28. Similarly, ∠YZU and ∠WXY are the exterior angle and the opposite interior angle, respectively.
By the property mentioned earlier, these angles are congruent.
The measure of ∠YZU is $2c_{∘}$ and the measure of ∠WXY is $a_{∘},$ which is equal to $96_{∘}.$ This information can be used to determine the value of c.
The value of c is 48.

$m∠VXY+m∠YXW=180_{∘}$

SubstituteII

$m∠VXY=84_{∘}$, $m∠YXW=a_{∘}$

$84_{∘}+a_{∘}=180_{∘}$

SubEqn

$LHS−84_{∘}=RHS−84_{∘}$

$a_{∘}=96_{∘}$

m∠VXY=m∠WZY

SubstituteII

$m∠VXY=84_{∘}$, $m∠WZY=3b_{∘}$

$84_{∘}=3b_{∘}$

DivEqn

$LHS/3=RHS/3$

$28_{∘}=b_{∘}$

RearrangeEqn

Rearrange equation

$b_{∘}=28_{∘}$

m∠YZU=m∠WXY

SubstituteII

$m∠YZU=2c_{∘}$, $m∠WXY=96_{∘}$

$2c_{∘}=96_{∘}$

DivEqn

$LHS/2=RHS/2$

$c_{∘}=48_{∘}$

Consider an inscribed quadrilateral EFGH. Draw a perpendicular bisector to each side of the polygon. **center** of the circle. This property is true for all cyclic quadrilaterals.

As can be observed, all the perpendicular bisectors intersect at the

It is worth mentioning that not only quadrilaterals can be inscribed in a circle. There can also be inscribed polygons with a different number of sides. Stonehenge is a real-world example of an inscribed polygon. Unfortunately, only some parts of it remain to this day.

However, when Stonehenge was built by ancient peoples about 5000 years ago, it had a cyclic polygon structure, as illustrated on the diagram below.

To sum up, inscribed quadrilaterals and polygons are not only interesting geometric objects — they can be seen and applied in real life. {{ 'mldesktop-placeholder-grade' | message }} {{ article.displayTitle }}!

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