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# Parallel Lines and Angles - PowerPoint PPT Presentation

A. B. Parallel Lines and Angles. Definition Two different given lines L 1 and L 2 on a plane are said to be parallel if they will never intersect each other no matter how far they are extended. . Definition Two angles are called vertical angles if they are opposite to each other

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B

Parallel Lines and Angles

Definition

Two different given lines L1 and L2 on a plane are said to be parallel if they will never intersect each other no matter how far they are extended.

Definition

Two angles are called vertical angles if they are opposite to each other

and are formed by a pair of intersecting lines.

Theorem

Any pair of vertical angles are always congruent.

L1

L2

T

Parallel Lines and Angles

Definition

Given two line L1 and L2 (not necessarily parallel) on the plane, a third line T is called a transversal of L1 and L2 if it intersects these two lines.

c

L1

L2

T

• Definitions

• Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T be a transversal.

• a and form a pair of corresponding angles.

• c and  form a pair of corresponding angles etc.

d

L1

L2

T

• Definitions

• Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T be a transversal.

• c and form a pair of alternate interior angles.

• d and  form a pair of alternate interior angles.

L1

L2

T

• Definitions

• Let L1 and L2 be two lines (not necessarily parallel) on the plane, and T be a transversal.

• a and form a pair of alternate exterior angles.

• b and  form a pair of alternate exterior angles.

• Theorem

• Let L1 and L2 be two lines on the plane, and T be a transversal.

• If L1 and L2 are parallel, then

• any pair of corresponding angles are congruent,

• any pair of alternate interior angles are congruent,

• any pair of alternate exterior angles are congruent.

L1

L2

T

• Theorem

• Let L1 and L2 be two lines on the plane, and T be a transversal.

• if there is a pair of congruent corresponding angles, then L1 and L2 are parallel.

• if there is a pair of congruent alternate interior angles, then L1 and L2 are parallel.

• if there is a pair of congruent alternate exterior angles, then L1 and L2 are parallel.

L1

L2

T

Definition

Given two triangles ΔABC and ΔXYZ.

If AB is congruent to XY, A is congruent to X ,

BC is congruent to YZ, B is congruent to Y , CA is congruent to ZX, C is congruent to Z

then we say that ΔABC is congruent to ΔXYZ, and we write

Y

B

Z

X

C

A

Side-Angle-Side Principle

Given two triangles ΔABC and ΔXYZ.

If AB is congruent to XY

B is congruent to Y

BC is congruent to YZ

then ΔABC is congruent to ΔXYZ

Z

B

Y

C

A

X

Angle-Side-Angle Principle

Given two triangles ΔABC and ΔXYZ.

If A is congruent to  X

AC is congruent to XZ

C is congruent to Z

then ΔABC is congruent to ΔXYZ

Z

((

B

Y

((

C

(

)

A

X

Side-Side-Side Principle

Given two triangles ΔABC and ΔXYZ.

If AB is congruent to XY

BC is congruent to YZ

CA is congruent to ZX

then ΔABC is congruent to ΔXYZ

Z

B

Y

C

A

X

If ΔABC is congruent to ΔXYZ , then

AB is congruent to XY

BC is congruent to YZ

CA is congruent to ZX

and

A is congruent to  X

B is congruent to  Y

C is congruent to  Z

In short, corresponding parts of congruent triangles are congruent.

D

B

E

C

Example 14.5

Show that the diagonals in a kite is perpendicular to each other.

Recall that a kite is a quadrilateral with 2 pairs of congruent adjacent sides. In particular for the following figure, AB = AD and CB = CD.

2

We first need to show that ΔADC and ΔABC are congruent.

This is true because AD = ABDC = BC

AC = AC

and we have the SSS congruence principle.

(b/c it is a kite)

(b/c it is a kite)

(b/c they are the same side)

A

D

B

Therefore, (click)

C

1 is congruent to  2

D

B

E

C

Now we only considerΔADE and ΔABE.

They should be congruent becauseAD = AB

 1 =  2AE = AE

Hence SAS principle applies.

1

2

 AED is congruent to  AEB, and they both add up to 180, hence each one is 90.

Definition

Given ΔABC and ΔXYZ.

If A is congruent to  X

 B is congruent to  Y

C is congruent to Z

and AB : XY = BC : YZ = CA : ZX

then we say that ΔABC is similar to ΔXYZ, and the notation is

ΔABC ~ ΔXYZ

Y

B

X

Z

A

C

SSS similarity principle

Given ΔABC and ΔXYZ.

If AB : XY = BC : YZ = CA : ZX

then ΔABC is similar to ΔXYZ.

Y

Z

B

A

C

X

AAA similarity principle

Given ΔABC and ΔXYZ.

If A is congruent to  X

B is congruent to  Y

C is congruent to  Z

then ΔABC is similar to ΔXYZ

X

B

C

A

Z

Y

AA similarity principle

Given ΔABC and ΔXYZ.

If A is congruent to  X

B is congruent to  Y

then ΔABC is similar to ΔXYZ

(because the angle sum of a triangle is always 180o)

X

B

C

A

Z

Y

SAS similarity principle

Given ΔABC and ΔXYZ.

If AB : XY = BC : YZ and

B is congruent to Y

then ΔABC is similar to ΔXYZ

Y

B

C

A

Z

X

If the shadow of a tree is 37.5 m long, and the shadow of

a 1.5m student is 2.5 m long. How tall is the tree?

(assuming that they are all on level ground.)

47m

B

40o

58m

C

Indirect Measurements

What is the distance between the two points A and B on the rim of the pond?

How far is the boat from the point A on shore?

B

How do we measure angles?

72º

38º

36m

C

A

land

A large (3.5m) optical rangefinder mounted on the flying bridge of the USS Stewart (a Destroyer Escort)

A Transit bridge of the USS Stewart (a Destroyer Escort)

is a surveying instrument to measure horizontal angles.

Glass reticle on both models has stadia lines for measuring distance. Stadia ratio 1:100

Measure the height of Devils Tower in Wyoming bridge of the USS Stewart (a Destroyer Escort)

The first 'proper' ascent was in 1937 when some of America's best climbers took on the project. The Weissner Route was the result, a 5.7 (decent VS) classic on which he placed a single per runner on the crux pitch.

A bizarre incident took place only a couple of years later, in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

His parachute descent went OK but on arrival he found that his rope had disappeared over the edge and he was well and truly stuck! There was only one solution, the first ascensionists were called upon to drive halfway across America to repeat their great feat and bring down the hapless Hopkins who had spent a cold and lonely week on his island in the sky.

Example 1. Use similar triangles to find the value of in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape. x.

x

7

Example 2. Use similar triangles to find the values of in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape. x and y.

x

y

R in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

4.5

x

S

Q

Example 2. Use similar triangles to find the values of x and y.

x

y

R in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

R

4.5

4.5

x

x

S

S

Q

Q

Example 2. Use similar triangles to find the values of x and y.

x

y

Example 2. Use similar triangles to find the values of in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape. x and y.

R

R

4.5

4.5

x

x

S

S

8

8

x

Q

Q

y

Example 3. Use similar triangles to find the values of in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape. x and y.

y

x

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

4

x

D

9

A

Example 3. Use similar triangles to find the values of x and y.

y

x

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

4

4

x

x

D

D

9

9

A

A

Example 3. Use similar triangles to find the values of x and y.

y

x

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

4

4

x

x

D

D

9

9

A

A

Example 3. Use similar triangles to find the values of x and y.

y

x

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

x

A

B

12

3

D

Example 4. Use similar triangles to find the value of x.

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

x

x

C

D

D

B

3

12

A

x

A

B

12

3

D

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

x

x

C

D

D

B

3

12

A

x

A

B

12

3

D

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

C

x

x

x

C

D

D

D

B

3

12

12

A

A

x

A

B

12

3

D

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

x

x

C

D

D

B

3

12

A

x

A

B

12

3

D

C in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

C

C

x

x

x

C

D

D

D

B

B

3

3

12

A

x

A

B

12

3

D

Constructions with Straight Edge and Compass in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

Remark

A straight edge is only used to draw line segments, and it should not have any marking on it. If a ruler is used, then all the markings should be ignored.

Reasons for in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape. not using a Protractor or Ruler

• We need to prepare for situations where some equipments are not available.

• Small protractors are not accurate enough for large scale projects.

• Many geometric constructions cannot be done by protractors but can be done by a compass and straight edge, such as finding the center of a circle.

What can the equipments do? in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

• A straight edge is infinitely long and is only used to draw line segments connecting given points or extend an existing line segment.

• A compass can be used to draw circles and circular arcs with radii that have already been constructed.

Basic Rules in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

Any construction should consist of only a finite number of steps. Only one step can be carried out at a time. Each construction must be exact. Approximation is not counted as a solution. In particular, a construction should not require drawing a line tangent to an existing circular arc because that will not provide accurate results.

A point must either be given in advance, or be constructed by the intersection of two line segments, or two arcs, or an arc and a line segment.

Basic Rules in 1941 when local air ace Charles George Hopkins decided to parachute onto the top of the tower to advertise his aerial show. He came prepared with a length of rope, a block and tackle as well as a sharpened axle from a Model T Ford to act as an anchor for his planned escape.

All compass and straightedge constructions consist of repeated application of five basic constructions using the points, lines and circles that have already been constructed or given. These are:

• Creating the line through two existing points

• Creating the circle through one point with centre at another point.

• Creating the point which is the intersection of two existing, non- parallel lines.

• Creating the one or two points in the intersection of a line and a circle.

• Creating the one or two points in the intersection of two circles.

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step I

Adjust the compass until the gap between the pencil and needle is more than half the length AB.

Move the needle tip of the compass to point A

(click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step II

Draw an approx. half circle as shown (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step II

Draw an approx. half circle as shown (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step III

Move the needle tip to point B. (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step III

Move the needle tip to point B. (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step III

Move the needle tip to point B. (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step III

Move the needle tip to point B. (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step IV

Draw an approx. half circle as shown (click)

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step IV

Draw an approx. half circle as shown

A

B

The following animation shows how to construct the perpendicular bisector (of a line segment) using just a straight edge and compass.

Step V

Use a ruler and a pencil to draw a line connecting the two intersections of the circular arcs.

This will be the perpendicular bisector of AB.

A

B

Standard examples perpendicular bisector (of a line segment) using just a straight edge and compass.

• Construction of an equilateral triangle, hence the construction of a 60o angles

• Construction of a right angle.

• Bisect a given angle.

• Construction of a square.

• Construction of a line segment parallel to a given line segment.

• Divide any given line segment into any given number of congruent parts.

• Locating the exact center of a given circle.

• Given a unit segment and any positive integer n, construction of a line segment whose length is n units.

• Given a unit segment and any positive real number x, construction of a segment whose length is sqrt of x times the unit segment.

• Given a unit segment and any positive real number x, construction of a segment whose length is 1/x times the unit segment.

Important Theorem perpendicular bisector (of a line segment) using just a straight edge and compass.

• A regular n-sided polygon is constructible if and only ifone of the following is true.

• n is a power of two (but bigger than 2), or

• n is an odd prime from the list 3, 5, 17, … (where the general form is 22k + 1 for some whole number k.)

• n is of the form

• 2mp1p2…pr

• where m is a whole number and pi’s are distinct odd primes of the above form.

Examples:

A regular polygon with 30 sides is constructible, but one with 35 sides will not be constructible. A regular polygon of 36 sides is also not constructible.

What kind of regular polygons are constructible? perpendicular bisector (of a line segment) using just a straight edge and compass.

• Any equilateral triangle is constructible.

• Any square is constructible.

• Any regular pentagon is constructible. (see lab 20)

• Any regular hexagon is constructible.

• Is a regular heptagon constructible? Why?

• Is a regular octagon constructible? Why?

No, because 7 is a prime not on the list of the previous theorem.

Yes, because 8 is a power of 2, and the previous theorem says

that it can be done.

What kind of regular polygons are constructible? perpendicular bisector (of a line segment) using just a straight edge and compass.

• Is a regular nonagon constructible? Why?

• Is a regular decagon constructible? Why?

• Is a regular 11-gon constructible? Why?

• Is a regular 12-gon constructible? Why?

• Is a regular 13-gon constructible? Why?

• Is a regular 14-gon constructible? Why?

No, b/c 9 = 3×3, the 3 is repeating.

Yes, b/c 10 = 2 ×5, and 5 is on the list.

No, b/c 11 is a prime but not on the list.

Yes, b/c 12 = 2×2×3 and 3 is on the list.

No, b/c 13 is a prime but not on the list.

No, b/c 14 = 2×7 but 7 is an odd prime not on the list.

Constructible Angles perpendicular bisector (of a line segment) using just a straight edge and compass.

We know that angles with measures of 90°, 60°, 45°, 30° are all constructible by straight edge and compass (see Lab 20).

1. Is it possible to construct a 75° angle (with straight edge and compass)?

Yes, b/c 75° = 45° + 30°

2. Is it possible to construct a 15° angle (with straight edge and compass)?

Yes, b/c 15° = 45° – 30°, and 15° = 30° ÷ 2

3. Is it possible to construct a 72° angle (with straight edge and compass)?

Yes, b/c a regular pentagon is constructible and its central angle has measure equal to 72°.

central angle

center

Central angle = 360° ÷ 5 = 72°

4. Is it possible to construct a 3° angle (with straight edge and compass)?

Yes, b/c 75° – 72° =3°

θ edge and compass)?

5. Is it possible to construct a 40° angle (with straight edge and compass)?

No, b/c the central angle θof a regular nonagon is exactly 360° ÷ 9 = 40°, and from a previous result, we know that a regular nonagon is not constructible.

6. Is it possible to construct a 1° angle (with straight edge and compass)?

No, b/c otherwise we can construct a 40° angle by repeated addition of angles.

• Conclusion edge and compass)?

• The smallest constructible angle with positive integer degree measure is a 3° angle.

• If n is a positive integer, and angle A has measure n°, then angle A is constructible if and only if n is a multiple of 3.

Remark:Conclusion II above does not apply to angles with non-integer degree measures. For instance, a 7.5° angle is constructible, but 7.5 is not a multiple of 3.

Example: Can we construct an 14° angle with just straight edge and compass?

2 edge and compass)?

3

What cannot be done?

• Trisect an angle whose measure is not 3 times of an angle which is already constructible by another method, for instance, 90o can be trisected because a 30o angle is constructible by bisecting a 60o angle.

• Square a circle: i.e. construct a square with the same area as a given circle.(consequently, circling a square is also impossible.)

• Construct a line segment whose length is

times a given unit line segment.

Enrichment edge and compass)?

Rectangle to Square Dissection Problem

Challenge: Can we dissect an arbitrary rectangle into a finite number of pieces and then rearrange those pieces into a square with exactly the same area?

The answers is yes, and it requires only 3 pieces if its length is no more than twice of its width. The following animation shows how this is done.

Step 1

Draw a square with exactly the same area as shown. (click)

A

Step 2

Construct the line segment AB. (click)

Click to see step 3

B

Rectangle to Square Dissection Problem edge and compass)?

Challenge: Can we dissect an arbitrary rectangle into a finite number of pieces and then rearrange those pieces into a square with exactly the same area?

The answers is yes, and it requires only 3 pieces if its length is no more than 4 times of its width. The following animation shows how this is done.

Step 1

Draw a square with exactly the same area as shown. (click)

Step 3

Slide the triangles up. (click)

Step 2

Construct the line segment AB. (click)

If the length is more than 4 times of the width, we first cut the rectangle into two congruent shorter pieces and then rearrange them into a rectangle with shorter length. (click to see animation)

Enrichment cut the rectangle into two congruent shorter pieces and then rearrange them into a rectangle with shorter length. (click to see animation)

Tarski’s circle-squaring problem

posed by Alfred Tarski in 1925, is to take a circle (including its interior) in the plane, cut it into finitely many pieces, and reassemble the pieces so as to get a square of exact area.

(click to see what will happen if we put the circle on top of the square.)

A partial solution was given by Miklos Laczkovich in 1990; whose decomposition consists of about 1050 different pieces, but the pieces do not have nice boundaries (i.e. not Jordan curves) and hence cannot be cut out with scissors.

It might be possible that some day, some one will give a better solution to this problem.

A regular 17-gon. whose decomposition consists of about 10

α