Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

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Section 4.3 Section 4.3 -A Right Angle Theorem -A Right Angle Theorem Michael Smertz Michael Smertz H Geometry- 8 H Geometry- 8 30 May 2008 30 May 2008

Transcript of Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Page 1: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Section 4.3Section 4.3-A Right Angle Theorem-A Right Angle Theorem

Section 4.3Section 4.3-A Right Angle Theorem-A Right Angle Theorem

Michael SmertzMichael SmertzH Geometry- 8H Geometry- 830 May 200830 May 2008

Page 2: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

The TheoremThe Theorem • In order to prove that lines are perpendicular, you must first prove that they form right angles.

• For this reason, it is necessary to know the following theorem:– Theorem 23: If two angles are both

supplementary and congruent, then they are right angles.

Page 3: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

How The Theorem Works

• Most of the problems dealing with this theorem will be proofs.

• Here is how you would use it in a proof if you were given the diagram at right. It is given that L1 is congruent to L2 and you must prove both angles are right angles.

1. Since L1 and L2 form a straight line, then they are supplementary.2. Then, since the angles are congruent, you know that each must equal 90°.3. Therefore, you now know that both of the angles are supplementary and congruent. You can now use the theorem that “If two angles are both supplementary and congruent, then they are right angles.”

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NOTE: You can now assume that whenever two angles form a straight line, they are supplementary.

Page 4: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Sample Problems• Problem #1

– Given: • Ray CD bisects LACB• Segment AC is

congruent to segment CB

– Prove:• LCDA and LCDB are

right angles

Solution

AB

C

D

1. Ray CD bisects LACB

2. AC congruent to CB

3. LACD congruent LBCD

4. CD congruent CD

5. ∆ACD congruent ∆BCD

6. LCDA congruent LCDB

7. LCDA and LCDB are right angles

1. Given

2. Given

3. If a ray bisects an angle, then it divides the angle into 2 congruent angles

4. Reflexive

5. SAS (2.3.4)

6. CPCTC

7. If two angles are both supplementary and congruent, then they are right angles.

Page 5: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Sample Problems

• Problem #2– Given:

• DA congruent to DC

• AB congruent to BC

– Prove: • DB altitude of AC

• Solution

D

A

B

C

E

1. DA congruent

to DC

2. AB congruent to BC

3. DB congruent to DB

4. ∆DAB congruent to ∆DCB

5. LABE congruent to LCBE

6. EB congruent to EB

7. ∆ABE congruent to ∆CBE

8. LAEB congruent to LCEB

9. LAEB and LCEB rt. L’s

10. DB alt. of AC

1. Given

2. Given

3. Reflexive

4. SSS (1,2,3)

5. CPCTC

6. Reflexive

7. SAS (2,5,6)

8. CPCTC

9. If two angles are both supplementary and congruent, then they are rt. L’s

10. An altitude of a ∆ forms right angles with the side to which it is drawn.Note: This is a detour problem.

Page 6: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Sample Problems

• Problem #3– If squares A and C are folded

across the dotted segments onto B, find the area of B that will not be covered by either square.

• Solution– In order to solve this

problem, you first have to find that the top part of B is eight. Then, fold over squares A and C to get the top part of B to be 4. Next, you know that the side of B will be two because A is a square when it is folded over. Lastly, you multiply two and four to find the area of B that will not covered by either square. The final answer is eight.

A B C

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2 2

Page 7: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Practice Problem #1

• Given:– P– AB congruent to BC

• Prove:– LDBC and LDBA are

right angles

A B C

D

Page 8: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Practice Problem #2M

O3x+14

2x+22

Is M perpendicular to O? Justify your

answer.

Page 9: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Practice Problem #3

• Given:– XY congruent to

XZ– XQ bisects LYXZ

• Prove:– XQ is

perpendicular to YZ

X

YQ Z

Page 10: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Practice Problem #4

• A diameter of a circle has endpoints with coordinates (1,6) and (5,8). Find the coordinates of the center of the circle.

(1,6)

(5,8)

Page 11: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Answer Sheet

Practice Problem #1

1. P, AB congruent to BC

2. Draw DC

3. AD congruent to DC

4. DB congruent to DB

5. ∆ADB congruent to ∆CDB

6. LDBA congruent to LDBC

7. LDBC and LDBA are right angles

1. Given

2. Two points determine a segment

3. All radii of a circle are congruent

4. Reflexive

5. SSS (1,3,4)

6. CPCTC

7. If two angles are both supplementary and congruent, then they are rt. L’s

Page 12: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Answer Sheet

Practice Problem #2

• YES 3x+14=2x+22

• X=8• 38=38

– This means the angles are congruent. Theorem #23 states, “If two angles are both supplementary and congruent, then they are rt. L’s.” The answer is yes because right angles are formed by perpendicular lines.

Page 13: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Answer SheetPractice Problem #3

1. XY congruent to XZ and XQ bisects LYXZ

2. LYXQ congruent to LZXQ

3. LY congruent to LZ

4. ∆YXQ congruent to ∆ ZXQ

5. LXQY congruent to LXQZ

6. LXQY and LXQZ are rt. L’s

1. Given

2. If a ray bisects an L, then it divides the L into 2 congruent L’s

3. If sides, then angles

4. ASA (1,2,3)

5. CPCTC

6. If two angles are both supplementary and congruent, then they are rt. L’s

7. Rt. L’s are formed by perpendicular lines

7. XQ is perpendicular to YZ

Page 14: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Answer SheetPractice Problem #4

1+5

2And

6+82

Answer: (3,7)

Page 15: Section 4.3 -A Right Angle Theorem Michael Smertz H Geometry- 8 30 May 2008.

Works Cited

"Chapter 2 Notes." 18 Oct. 2007. 29 May 2008

<home.cvc.org/math/dgeom/Chapter_2_notes/2_8_2.pdf>.

Rhoad, Richard, George Milauskas, and Robert Whipple.

Geometry for Enjoyment and Challenge. New ed. Evanston:

McDougal, Littell & Company, 1997. 180-183.