mathematics


Semester 2 Lore Project: Proofs

Student calculate:

Overview: The intention of this design is to make-trial-of a few geometric theorems. The design is

divided into two activities, each requiring one demonstration. The demonstrations get recount to topics that you'll

cover in advenient chapters. The highest demonstration get be a three-part, two-column demonstration. The instant get be

a behalf demonstration.


Your online passagebook get be an inestimable regard for this design. In each spectre,

the lore individuality get substantiate the behalf of your passagebook most pertinent to the required

proof.


Instructions: To perfect the design, you'll content in the passage boxes (for stance,      ) after a while

your answers. This rasp is set up as a reader-enabled contrive. This media you can merely enter

content into the required fields. To saunter through the rasp, hit tab or click in the passage boxes to

enter your answers. Hitting tab get assume you to each of the fields you want to perfect for the

project. Often, precedently entering your answers in the passage boxes, you'll want to do some product on

dally brochure.


Once you keep contented in all your answers, elect “Save As” from the Rasp menu. Learn your

scholar calculate in the rasp indicate precedently you upload your assignment to Penn Foster. For

example, the rasp you downloaded the rasp indicated scholar-number_0236B12S.pdf. When the

window appears to "Save As," learn your scholar calculate in the rasp indicate

(12345678_0236B12S.pdf), where 12345678 is your eight-digit scholar calculate).

Course address and calculate: MA02B01

Assignment calculate: 0236B12S

Page 1 of 4

Activity 1: Demonstration of the SSS Consonance Theorem


Theorem 8.3.2: If the three sides in one triintention are proportional to the three sides in another

triangle, then the triangles are congruous.


Setup: On dally brochure, describe two triangles after a while one larger than the other and the sides of one

triintention proportional to the other. Label the larger triintention ABC and the smaller triintention DEF so

that




Given: The sides of triintention ABC are proportional to the sides of triintention DEF so that





Prove: Triintention ABC is congruous to triintention DEF.


Research: In your online passagebook, examine Chapter 8 to learn properties of congruousity. If

necessary, resurvey reasoning and demonstration in Chapter 2, properties of correspondent lines in Chapter 3, and

triintention congruence in Chapter 4. To perfect this demonstration, you may use any specification, assume,

or theorem in your online passagebook on or precedently page 517.


Statements

1. Limb GH is correspondent to limb BC.

2. Limb AB and AC are to

segments GH and BC.

3. Intention AGH is to intention ABC,

and intention AHG is congruent to intention ACB.

4. Triintention AGH is to triintention ABC.


Reasons

1. By

2. Specification of a


3. Angles Postulate

4. AA Property

Page 2 of 4

Proof:

Part 1: Compose limb GH in triintention ABC so that G is betwixt A and B, AG = DE, and

limb GH is correspondent to limb BC. (Hint: You should substantially do this on your setup

figure.) Show that triintention AGH is congruous to triintention ABC.

Part 2: Show that triintention AGH is congruent to triintention DEF.

Page 3 of 4


Statements

1. Triintention AGH is to triintention ABC.

2. The sides of triintention AGH are

to the sides of triintention ABC.

3. The sides of triintention ABC are proportional to

the sides of triintention DEF.

4. The sides of triintention AGH are

to the sides of triintention DEF.

5.



6. AG = DE

7. GH = EF and HA = FD

8. Triintention AGH is to triintention DEF.


Reasons

1. Fruit from Part 1


2. Polygon Postulate

3.

4. Property

5. Specification of sides


6. By

7. Transitive Property

8. Congruency Postulate

Part 3: Show the required fruit.


Statements

1. Triintention AGH is to triintention DEF.

2. Intention AGH is to intention DEF, and

intention GAH is congruent to intention EDF.

3. Intention AGH is congruent to intention ABC, and

intention AHG is to intention ACB.

4. Intention ABC is to intention DEF, and

intention ACD is congruent to intention EDF.

5. Triintention ABC is to triintention DEF.


Reasons

1. Fruit from Part 2

2.

3. Repeat of declaration shown in Part 1

4. Property

5. AA Property

Note: You can make-trial-of the SAS Consonance Theorem in relish mould.

Activity 2: Demonstration of the Converse of the Chords and Arcs Theorem


Theorem 9.1.6: In a divergence or in congruent divergences, the chords of congruent arcs are congruent.


Setup: On dally brochure, compose congruent divergences after a while centers at P and M. Then compose

congruent arcs QR on divergence P and NO on divergence M. Finally, describe triangles PQR and MNO.


Research: In your online passagebook, examine Chapter 9 to learn the properties of arcs and

circles in open. If essential, resurvey reasoning and demonstration in Chapter 2 and triintention congruence

in Chapter 4. To perfect this demonstration, you may use any assume or theorem on or precedently 568 in

your online passagebook.


Proof: Since QR and NO are , intention QPR is congruent to intention NMO by the

 of the step gauge of arcs.


Segments PQ, PR, MN, and MO are all of congruent divergences, so they are all .

In feature, limb PQ is to limb MN and limb PR is congruent to

 MO.


Therefore, triintention PQR is to triintention MNO by . Consequently, limb QR

is congruent to limb NO by , which make-trial-ofs that, in a

 or in congruent divergences, the of congruent are congruent.