## Math 9 Chapter 3 Lesson 7: Inscribed quadrilateral

## 1. Summary of theory

### 1.1. Concept

**Define**: A quadrilateral with four vertices lying on the same circle is called a cyclic quadrilateral (or inscribed quadrilateral).

For example, the quadrilateral \(ABCD\) has four vertices \(A,B,C,D\) lying on the same circle, so \(ABCD\) is called an inscribed quadrilateral.

### 1.2. Theorem:

In an inscribed quadrilateral, the sum of the measures of two opposite angles is 180 .^{0}

\(ABCD\) is an inscribed quadrilateral so we have \(\widehat{A}+\widehat{C}=\widehat{B}+\widehat{D}=180^0\)

### 1.3. Island theorem

If a quadrilateral has the sum of the measures of two opposite angles equal to 180^{0} then the quadrilateral is inscribed in the circle

Specifically in the image above, if there is \(\widehat{A}+\widehat{C}=180^0\) or \(\widehat{B}+\widehat{D}=180^0\) then the quadrilateral is \(ABCD\) is inscribed in a circle.

### 1.4. Some signs to recognize inscribed quadrilaterals

– A quadrilateral whose sum of two opposite angles is equal to \(180^\circ \).

A quadrilateral whose exterior angle at a vertex is equal to the interior angle at the vertex for that vertex.

– A quadrilateral with four vertices equidistant from a point (which can be determined). That point is the center of the circumcircle of the quadrilateral.

– Quadrilateral with two adjacent vertices look at the side containing the remaining two vertices under the same angle \(\alpha \).

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **

a) Draw a circle with center O and then draw a quadrilateral whose vertices all lie on that circle.

b) Draw a circle with center I and then draw a quadrilateral whose three vertices lie on the circle and the fourth vertex is not.

__Solution guide__

**Verse 2:** Calculate the measure of the angles of the quadrilateral \(ABCD\), knowing that \(\widehat{DCx}=130^0\)

__Solution guide__

We have \(\widehat{DCB}=180^0-\widehat{DCx}=180^0-130^0=50^0\), so \(\widehat{DAB}=180^0-\widehat {DCB}=180^0-50^0=130^0\)

Again \(\widehat{DCx}\) is the outer corner of \(\bigtriangleup ECB\) so \(\widehat{DCx}=\widehat{E}+\widehat{B}\Rightarrow \widehat{B}= \widehat{DCx}-\widehat{E}=130^0-30^0=100^0\)

From this it follows that \(\widehat{ADC}=180^0-\widehat{ABC}=180^0-100^0=80^0\)

### 2.2. Advanced exercises

**Question 1: **See Figure 45. Let’s prove the theorem above.

__Solution guide__

Considering the circle \((O)\) we have:

\(\widehat {BAD} = \dfrac{1}{2}sđ\,\overparen {BCD}\) (inscribed angle intercepting arc \(BCD\))

\(\widehat {BCD} = \dfrac{1}{2}sđ\,\overparen {BAD}\) (inscribed angle intercept \(BAD\))

So \(\widehat {BAD} + \widehat {BCD} = \dfrac{1}{2}sđ\,\overparen {BCD} + \dfrac{1}{2}sđ\,\overparen {BAD} = \dfrac{{sđ\,\overparen {BAD} + sd\,\overparen {BCD}}}{2}\) \( = \dfrac{{360^\circ }}{2} = 180^\circ . \)

So \(\widehat {BAD} + \widehat {BCD} = 180^\circ \) .

So in an inscribed quadrilateral, the sum of the measures of two opposite angles is \(180^0\).

**Verse 2:** Let \(ABC\) triangle right at \(A,(AB .)

a) \(CI\) is the bisector of \(\widehat{BCD}\)

b) \(DA\) is tangent to \((O)\).

__Solution guide__

a) We have \(\widehat{IDC}=90^0\) (inscribed angle intercepting diameter)

So \(\widehat{BAC}=\widehat{BDC}=90^0\) deduces the inscribed quadrilateral \(ABCD\)

so \(\widehat{ACD}=\widehat{ABD}\) which according to the title \(\widehat{ABD}=\widehat{ACB}\) should \(\widehat{ACD}=\widehat{ACB} \) or \(CI\) is the bisector of \(\widehat{BCD}\) (dpcm)

b) The inscribed quadrilateral \(ABCD\) should \(\widehat{ADB}=\widehat{ACB}\) and \(\widehat{ACD}=\widehat{ACB}\) should \(\widehat{ADB) }=\widehat{ACD}\)

From this it follows that \(DA\) is a tangent to \((O)\).

## 3. Practice

### 3.1. Essay exercises

**Question 1: **On a circle with center \(O\) there is an arc \(AB\) and \(S\) is the midpoint of that arc. On the string \(AB\) take two points \(E\) and \(H.\) The lines \(SH\) and \(SE\) intersect the circle respectively at \(C\) and \ (D.\) Prove that \(EHCD\) is a cyclic quadrilateral.

**Verse 2: **Given a triangle \(ABC.\) The interior bisectors of \(\widehat B\) and \(\widehat C\) intersect at \(S,\) the exterior bisectors of \(\widehat B \) and \(\widehat C\) intersect at \(E.\) Prove that \(BSCE\) is a cyclic quadrilateral.

**Question 3: **Let an isosceles triangle \(ABC\) with bases \(BC\) and \(\widehat A = {20^0}\). On the shore half-plane \(AB\) does not contain the point \(C\) take the point \(D\) such that \(DA = DB\) and \(\widehat {DAB} = {40^0}\) . Let \(E\) be the intersection of \(AB\) and \(CD.\)

\(a)\) Prove that \(ACBD\) is a cyclic quadrilateral

\(b)\) Calculates \(\widehat {AED}\)

**Question 4: **Given three circles passing through the same point \(P.\) Call the other intersections \(P\) of two of the three circles \(A, B, C.\) From a point \(D\) ) (different from \(P\)) on the circle \((PBC)\) draw rays \(DB, DC\) that intersect the circles \((PAB)\) and \((PAC)\) times turns at \(M, N.\) Prove that the three points \(M, A, N\) are collinear.

### 3.2. Multiple choice exercises

**Question 1: **Let ABC be a triangle inscribed in circle (O; R), altitude AH. Know that AB=12cm, AC=20cm, AH=10m. The length of the radius of the circle is:

A. 9cm

B. 10cm

C. 11cm

D. 12cm

**Verse 2:** Given a circle (O;6cm) with diameter AD. The chord BC of the circle intersects AD at I (I lies between A and O). Know IB=4cm, IC=5cm. The AI length is:

A. 5cm

B. 4cm

C. 3cm

D. 2cm

**Question 3: **Which of the following statements is incorrect:

A. An inscribed quadrilateral is a quadrilateral with four vertices lying on the same circle.

B. If a quadrilateral whose sum of opposite angles is 180^{0} then the quadrilateral is inscribed in the circle.

C. In a cyclic quadrilateral, the sum of the measures of two opposite angles is 180^{0}.

D. Any quadrilateral can always be inscribed in a circle.

**Question 4: **Given a point A outside the circle (O), draw the line ABC to the circle and two tangents AE,AF to the circle. Let H be the intersection of AO and EF. Which of these following statements is wrong:

A. \(AO\perp EF\)

B. HE is the bisector of \(\widehat{BHC}\)

C. \(AB.AC=AO^2\)

D. Inscribed quadrilateral BHOC.

## 4. Conclusion

Through this lesson, you will learn some key topics as follows:

- Know the definitions and properties of angles of inscribed quadrilaterals; know that there are quadrilaterals that can be inscribed and there are quadrilaterals that cannot inscribe any circles. Know the conditions for a quadrilateral to be inscribed (necessary and sufficient conditions).
- Using properties of inscribed quadrilaterals and identifying signs of inscribed quadrilaterals in math and practice.

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