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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*.
### Rule

## Inscribed Quadrilateral Theorem

### 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 $2$

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.

$m∠A+m∠C=180_{∘}$

$m∠B+m∠D=180_{∘}$

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 $ABCE.$

$m∠AEC+m∠B=180_{∘} $

It was assumed that $ABCD$ has supplementary opposite angles. Therefore, $∠D$ and $∠B$ are supplementary angles.
$m∠D+m∠B=180_{∘} $

By the Transitive Property of Equality, the above equations imply that $∠AEC$ and $∠D$ have equal measures.
$m∠AEC+m∠B=180_{∘}m∠D+m∠B=180_{∘} ⇓m∠AEC=m∠D $

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 $∠D.$
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_{∘}.$$m∠ADE+m∠ADC=180_{∘} $

Also, by the Inscribed Quadrilateral Theorem, the opposite angles $∠ABC$ and $∠ADC$ are also supplementary.
Therefore, the following relation is true.
$m∠ABC+m∠ADC=180_{∘} $

After analyzing the equations, $∠ABC$ and $∠ADE$ show to be supplementary to the same angle $∠ADC.$ Therefore, by the Congruent Supplements Theorem, they are congruent. $∠ABC≅∠ADE$

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.
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.
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 value of $c$ is $48.$

$m∠VXY+m∠YXW=180_{∘} $

By substituting $84_{∘}$ for $m∠VXY$ and $a_{∘}$ for $m∠YXW$ into the equation, the value of $a$ can be calculated.
$m∠VXY+m∠YXW=180_{∘}$

SubstituteII

$m∠VXY=84_{∘}$, $m∠YXW=a_{∘}$

$84_{∘}+a_{∘}=180_{∘}$

SubEqn

$LHS−84_{∘}=RHS−84_{∘}$

$a_{∘}=96_{∘}$

$∠VXY≅∠WZY⇓m∠VXY=m∠WZY $

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.
$m∠VXY=m∠WZY$

SubstituteII

$m∠VXY=84_{∘}$, $m∠WZY=3b_{∘}$

$84_{∘}=3b_{∘}$

DivEqn

$LHS/3=RHS/3$

$28_{∘}=b_{∘}$

RearrangeEqn

Rearrange equation

$b_{∘}=28_{∘}$

$∠YZU≅∠WXY⇓m∠YZU=m∠WXY $

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 $c.$
$m∠YZU=m∠WXY$

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.