## Math 9 Chapter 3 Lesson 6: The arc containing the angle

## 1. Summary of theory

### 1.1. How to solve locus problem

To prove that the locus (set) of points M satisfying the property \(\tau\) is a certain shape \(H\), we have to prove two parts:

**Pros:** Every point with property \(\tau\) is in the shape \(H\).

**Island part:** Every point in the shape \(H\) has the property \(\tau\).

**Conclusion: **The locus (or set) of points M with property \(\tau\) is the shape \(H\)

**Comment: **A locus problem is easier to deal with when we can predict the figure \(H\) before we begin the proof.

### 1.2. The locus problem “Arc contains angle”

Given the given line segment \(AB\) and angle \(\alpha(0^0<\alpha<180^0)\) then the locus of points \(M\) satisfies \(\widehat{AMB}= \alpha\) are two arcs containing the angle \(\alpha\) built on the segment \(AB\)

**Attention:**

– The two arcs containing the angle \(\alpha\) above are two arcs that are symmetrical to each other through \(AB\)

– Two points \(A,B\) are considered to be in the locus

– In case \(\alpha=90^0\), the locus above is two semicircles of diameter \(AB\)

Apply the arc containing the angle to prove that the four points are on the same circle: If a quadrilateral with two adjacent vertices looks at the side containing the other two vertices under an angle \(\alpha\), then the four vertices of that quadrilateral have the same angle. belongs to a circle.

### 1.3. How to draw arc containing angle α

– Draw the perpendicular bisector d of the line segment AB.

– Draw the ray Ax making an angle α with AB.

– Draw a line Ay perpendicular to Ax. Let O be the intersection of Ay with d.

– Draw arc AmB, center O, radius OA such that this arc lies in the half-plane of edge AB that does not contain Ax rays.

– The arc AmB drawn as above is an arc containing angle α.

## 2. Illustrated exercise

### 2.1. Basic exercises

**Question 1: **Given the line segment \(CD\).

a) Draw three points \(N_1;N_2;N_3\) such that \( \widehat {CN_1D}=\widehat {CN_2D}=\widehat {CN_3D}=90^0\)

b) Prove that the points \(N_1;N_2;N_3\) lie on the circle of diameter \(CD.\)

**Solution guide**

a) Draw a picture.

b) Let \(I\) be the midpoint of the side \(CD.\)

Since the triangle \(C{N_1}D\) is right-angled at \({N_1}\) \(I{N_1} = IC = ID = \dfrac{{CD}}{2}\)

Similar to two right triangles \(C{N_2}D;C{N_3}D\) we have \(I{N_2} = I{N_3} = IC = ID = \dfrac{{CD}}{2} \)

So \(I{N_1} = I{N_2} = I{N_3} = \dfrac{{CD}}{2}\) or \({N_1};{N_2};{N_3}\) belongs to the circle. glasses \(CD.\)

**Verse 2:** Given trapezoid ABCD (AB//CD), O is the intersection of the two diagonals. On ray OA take point M such that OM=OB. On ray OB take point M such that ON=OA. Prove that four points D,M,N,C lie on the same circle.

**Solution guide**

Consider two triangles \(\bigtriangleup AOB\) and \(\bigtriangleup NOM\) having \(\widehat{AOB}\) in common and OA=ON; OM=OB

so \(\bigtriangleup AOB=\bigtriangleup NOM\)(cgc)

deduce \(\widehat{BAO}=\widehat{MNO}\)

On the other hand, since AB//CD (trapezoid) should \(\widehat{BAO}=\widehat{DCO}\), from which it follows that \(\widehat{MNO}=\widehat{DCO}\)

Consider the quadrilateral DMNC with \(\widehat{MNO}=\widehat{DCO}\) and these two angles look at the side MD, so the four points D,M,N,C belong to the same circle.

### 2.2. Advanced exercises

**Question 1: **Given a fixed arc AB formed by radii OA,OB perpendicular to each other, point I moves on arc AB. On the ray OI take a point M such that OM is equal to the sum of the distances from I to OA and OB. Find locus of points M.

**Solution guide**

**Pros:** Draw \(IH\perp OA,IK\perp OB\), point M on OI with property OM=IH+IK (1)

Draw \(BE\perp OI\). We have \(\bigtriangleup OBE=\bigtriangleup OIK\) (hype – acute angle) so OE=OK=IH, BE=IK (2)

From (1) and (2) deduce OM=IH+IK=OE+BE and hence EM=EB

Infer that triangle EMB is right-angled at E, so \(\widehat{EMB}=45^0\). Point M looks at OB fixed at angle \(45^0\) so M moves on arc containing angle \(45^0\) built on OB.

On the other hand, since the point M lies only inside the right angle AOB, M moves only on the arc AmB, the part of the arc containing the angle \(45^0\) built on OB.

**Island part:** Take any point M on arc AmB. Draw \(BE\perp OM,IH\perp OA, IK\perp OB\) we will prove OM=IH+IK

Indeed, we do the opposite of the positive part

Since \(\widehat{OMB}=45^0\) triangle EMB is right-angled at E, deducing EM=EB

\(\bigtriangleup OBE=\bigtriangleup OIK\) (hype – acute angle) so OE=OK=IH, BE=IK. Hence EM=IK

So OM=OE+EM=IH+IK

**Conclusion: **The locus (set) of points M is the arc AmB, part of the arc containing the angle \(45^0\) built on the segment OB inside the right angle AOB.

**Verse 2: ** Given a semicircle (O) of diameter AB. C is a point on the semicircle. On radius OC take a point D such that OD is equal to the distance from C to AB.

**Solution guide**

**Pros: **Draw \(OP\perp AB\) with P in (O)

Consider \(\bigtriangleup OPD\) and \(\bigtriangleup COH\) with

OD=OH (assumption)

OP=OC (same semicircle radius)

\(\widehat{POD}=\widehat{OCH}\) (staggered in)

So \(\bigtriangleup OPD=\bigtriangleup {COH}\) (cgc) deduces \(\widehat{ODP}=90^0\)

On the other hand, we have fixed O, P, so D lies on the circle of diameter OP

**Island part: **Take any point D’ lying on the circle of diameter OP, ray OD’ intersecting (O) at C’. Lower the perpendicular C’H’ to AB. We will prove OD’=C’H’

Indeed, consider two right triangles OD’P and C’H’O with hypotenuse OP=OC’ and acute angle \(\widehat{POD’}=\widehat{OC’H’}\)(staggered) in)

So \(\bigtriangleup OD’P=\bigtriangleup C’H’O\) (hype – acute angle) infer OD’=CH’

**Conclusion:** The locus (set) of points D when C runs on a semicircle of diameter AB is a circle of diameter OP with P being the midpoint of arc AB.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let \(ABC\) triangle with side \(BC\) fixed and \(\widehat A = \alpha \) constant. Find the locus of intersection of the three interior bisectors of the triangle

**Verse 2:** Construct a triangle \(ABC,\) knowing \(BC = 3 cm,\) \(\widehat A = {45^o}\) and the median \(AM = 2.5 cm.\)

**Question 3: **Given a fixed diameter \(AB\) semicircle. \(C\) is the point on the semicircle, on the string \(AC\) extends take the point \(D\) such that \(CD = CB.\)

\(a)\) Find the locus of points \(D\) when \(C\) runs on the given semicircle.

\(b)\) On the ray \(CA\) take the point \(E\) such that \(CE = CB.\) Find the locus of the points \(E\) when \(C\) runs on the halfway line given round.

**Question 4:** Let a semicircle of diameter \(AB\) and \(C\) be a point on the semicircle. On the radius \(OC\) take the point \(D\) such that \(OD\) is equal to the distance \(CH\) from \(C\) to \(AB.\) Find locus of points \( D\) when \(C\) runs on the given semicircle.

### 3.2. Multiple choice exercises

**Question 1:** The locus of points M looks at line segment AB under an angle of 120^{0} to be:

A. A circle passes through two points A,B

B. A line parallel to AB

C. An arc containing an angle of 120^{0} built on two points A,B

D. Two arcs containing an angle of 120^{0} Symmetrical on two points A and B

**Verse 2: **Given a line d, a point C lies outside the line d and is 5 cm from d. The set of points on d at a distance of 6 cm from C is

A. Two points on d are 6cm . from C

B. No points

C. Is a straight line d

D. A point on d is about 6cm from C.

**Question 3: **Which of these following statements is wrong:

A. The set of points equidistant from the two endpoints of a line segment is the perpendicular bisector of that line

B. The set of points that lie outside a given angle and are equidistant from the two rays of that angle is the internal bisector drawn from the vertex of that angle.

C. The set of points that are separated from a given point is a circle with the center at the given point

D. Given the given line segment \(AB\) and angle \(\alpha(0^0<\alpha<180^0)\) then the locus of points \(M\) satisfies \(\widehat{AMB) }=\alpha\) are two arcs containing the angle \(\alpha\) built on the segment \(AB\)

**Question 4: **Let ABC be a right triangle at A, with side BC fixed. Let I be the intersection of the three internal bisectors. The locus of point I when A changes is:

A. Circle with diameter BC

B. Line parallel to BC

C. An arc containing an angle of 135^{0} built on segment BC (located on the same side of A relative to BC)

D. Two arcs containing an angle of 135^{0} built on segment BC

## 4. Conclusion

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

- The arc locus contains an angle, apply the pair of forward and reverse propositions of this locus to solve the problem.
- Use the correct term “arc containing an angle” to construct on a line segment and draw an arc containing an angle on a given line segment.
- Solve the locus problem consisting of two parts: forward, reverse and conclusion.

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