## Math 9 Chapter 3 Lesson 3: Inscribed angle

## 1. Summary of theory

### 1.1. Define

An inscribed angle is an angle whose vertex lies on a circle and whose sides contain two chords of that circle.

The arc inside the angle is called the intercepted arc.

The angle \(\widehat{BAC}\) is called the inscribed angle, the intercepted arc is \(BC\)

In a circle, the measure of the inscribed angle is half the measure of the intercepted arc.

In a circle:

– Congruent inscribed angles intercept congruent arcs

– Inscribed angles that intercept the same arc or intercept equal arcs are congruent

– An inscribed angle (less than or equal to 900) has a measure equal to half the measure of the central angle intercepting the same arc

The inscribed angle that intercepts a semicircle is a right angle

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **Based on the figure, calculate the measure of arc \(BD\) small

**Solution guide**

\(\bigtriangleup OAD\) is at \(O\) so \(\widehat{OAD}=\frac{180^0-150^0}{2}=15^0\), deduce \(\widehat {BAD}=30^0+15^0=45^0\)

Since \(\widehat{BAD}\) is an inscribed angle, sd\(\stackrel\frown{BD}=2.\widehat{BAD}=2.45^0=90^0\)

**Verse 2:** Calculate \(\widehat{MON}\) if the measure of minor arc XY of a circle with center B is 700

**Solution guide**

In the circle \((B)\) we have sd\(\stackrel\frown{XY}=70^0\Rightarrow \widehat{XBY}=70^0\)

In a circle \((O)\) then \(\widehat{MON}=2.\widehat{MBN}=2.70^0=140^0\)

**Question 3:** Given circle \((O)\) and string \(AB\). Draw \(OH\perp AB(H\in AB)\), \(OH\) to intersect the minor arc \(AB\) at \(N\). Knowing that \(HN=5,AB=10\sqrt{5}\). Calculate the radius of the circle \((O)\)

**Solution guide**

Draw the diameter \(NOM\). It is easy to prove that \(H\) is the midpoint of \(AB\) so \(AH=\frac{1}{2}.AB=\frac{1}{2}.10\sqrt{5}=5 \sqrt{5}\)

Applying the trigonometric equation for right triangle MAN with altitude AH we have \(MH.HN=AH^2\Rightarrow MH=\frac{AH^2}{NH}=\frac{(5\sqrt{5} )^2}{5}=25\)

Then \(MN=MH+HN=25+5=30\)

The radius of the circle \((O)\) is \(ON=\frac{MN}{2}=15\)

### 2.2. Advanced exercises

**Question 1:** Given a circle \((O;R)\) of diameter \(BC\) fixed. The point \(A\) moves on a circle other than \(B\) and \(C\). Draw the diameter \(AOD\). Locate the point \(A\) so that the area \(\bigtriangleup ABC\) reaches the maximum value, then \(\widehat{ADC}=?\)

**Solution guide**

Draw the altitude \(AH\) of \(\bigtriangleup ABC\).

\(\bigtriangleup AHO\) is square at \(H\) so \(AH\leq AO\) (equal sign occurs when \(H\equiv O\))

\(S_{ABC}=\frac{1}{2}AH.BC\leq \frac{1}{2}.AO.BC=\frac{1}{2}R.2R=R^2\) (equal sign occurs when \(H\equiv O\))

So the triangle area \(ABC\) reaches its maximum value when \(H\equiv O\), then \(A\) is the midpoint of \(\stackrel\frown{BC}\)

Derive \(\widehat{ADC}=45^0\)

**Verse 2: **Given a semicircle of diameter \(AB=2cm\), string \(CD//AB (C\in\stackrel\frown{AD})\). Find the lengths of the sides of a trapezoid \(ABCD\) whose perimeter is \(5cm\)

**Solution guide**

We have \(CD//AB\Rightarrow \stackrel\frown{AC}=\stackrel\frown{BD}\Rightarrow AC=BD\). It is easy to prove that \(ABDC\) is an isosceles trapezoid (because \(\widehat{CAB}=\widehat{DBA}\))

Set \(AC=BD=x\) \((x>0)\), the perimeter of the trapezoid is \(5cm\) so \(AB+BD+CD+AC=5\Rightarrow CD=3-2x\ )

Draw \(DN,CM\) perpendicular to \(AB\). We have \(NB=MA=\frac{AB-CD}{2}=\frac{2-(3-2x)}{2}=\frac{2x-1}{2}\)

\(\bigtriangleup DAB\) square at \(D\) has \(DN\perp AB\) so \(BD^2=BN.BA\Rightarrow x^2=\frac{2x-1}{2}. 2\Rightarrow x^2-2x-1=0\Rightarrow x=1\)

So \(AC=BD=1cm\) , so \(CD=3-2x=1 (cm)\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Given a circle with center O and radius 1.5cm. Draw a square ABCD with four vertices lying on the circle. Show how to draw

**Verse 2:** Given a circle (O) and a fixed point M not lying on the circle. Through M draw any sandline that intersects the circle at A and B. Prove that the product MA.MB is constant.

**Question 3: **In order to help the train to change from one track to another, one makes an arc-shaped track in the middle (right figure). Knowing the width of the track is AB ≈1,1m, BC ≈28, 4m.Calculate the radius OA=R of the arc track

**Question 4: **Let ABC be an equilateral triangle inscribed in circle (O) and M is a point of minor arc BC. On MA take point D such that MD = MB

a. What is the triangle MBD?

b. Compare two triangles BDA and BMC

c. Prove that MA = MB + MC

### 3.2. Multiple choice exercises

**Question 1: **Indicate which of the following statements is true:

A. Congruent inscribed angles intercept equal arcs.

B. The inscribed angle that intercepts a semicircle is a right angle.

C. Inscribed angles that intercept the same arc or intercept equal arcs are congruent.

D. In a circle, the measure of the inscribed angle is half the measure of the intercepted arc.

**Verse 2: **Let a circle (O) with diameter AB equal to 12cm. A line through A intersects the circle (O) at M and intersects the tangent to the circle at B at N. Let I be the midpoint of MN. Given that AI=13cm, the measure of line segment AM is:

A. 10cm

B. 6cm

C. 8cm

D. 12cm

**Question 3: **Based on the following figure, know AB and CD are the two chords of the circle (O), M is the midpoint of the minor arc AB. Which of these following statements is wrong:

A. \(\widehat{ADM}=\widehat{BDM}\)

B. \(\widehat{BCM}=\widehat{ABM}\)

C. \(AM=BM\)

D. If M is midpoint of major arc CD then \(\widehat{MDC}=\widehat{DCB}\)

**Question 4:** Based on the following figure, know that \(\widehat{AOB}=130^0,\widehat{ADO}=40^0\) and sd\(\stackrel\frown{CD} =30^0\). The measure of angle BAC is:

A. 50^{0}

B. 60^{0}

C. 70^{0}

D. 80^{0}

## 4. Conclusion

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

- Identify the inscribed angles on a circle and the defined statement about the inscribed angle.
- State and prove the theorem about the measure of an inscribed angle.
- Visually identify and prove the consequences of the above theorem.
- Classify the cases of inscribed angles.

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