What is the measure of angle d f e? triangle d f e. the exterior angle to angle f is 28 degrees.

Inhaltsverzeichnis Show

Taking the Burden out of Proofs
Proving Segment and Angle Relationships
Proving Relationships Between Lines
Two's Company. Three's a Triangle
Congruent Triangles
Smiliar Triangles
Opening Doors with Similar Triangles
Putting Quadrilaterals in the Forefront
Anatomy of a Circle
The Unit Circle and Trigonometry

This provides the answers and solutions for the Put Me in, Coach! exercise boxes, organized by sections.

Taking the Burden out of Proofs

Yes
Theorem 8.3: If two angles are complementary to the same angle, then these two angles are congruent.

A and B are complementary, and C and B are complementary.

Given: A and B are complementary, and C and B are complementary.

Prove: A ~= C.

	Statements	Reasons
1.	A and B are complementary, and C and B are complementary.	Given
2.	mA + mB = 90º , mC + mB = 90º	Definition of complementary
3.	mA = 90 º - mB, mC = 90º - mB	Subtraction property of equality
4.	mA = mC	Substitution (step 3)
5.	A ~= C	Definition of ~=

Proving Segment and Angle Relationships

If E is between D and F, then DE = DF EF.

E is between D and F.

Given: E is between D and F

Prove: DE = DF EF.

	Statements	Reasons
1.	E is between D and F	Given
2.	D, E, and F are collinear points, and E is on ¯DF	Definition of between
3.	DE + EF = DF	Segment Addition Postulate
4.	DE = DF EF	Subtraction property of equality

2. If BD divides ABC into two angles, ABD and DBC, then mABC = mABC - mDBC.

BD divides ABC into two angles, ABD and DBC.

Given: BD divides ABC into two angles, ABD and DBC

Prove: mABD = mABC - mDBC.

	Statements	Reasons
1.	BD divides ABC into two angles, ABD and DBC	Given
2.	mABD + mDBC = mABC	Angle Addition Postulate
3.	mABD = mABC - mDBC	Subtraction property of equality

3. The angle bisector of an angle is unique.

ABC with two angle bisectors: BD and BE.

Given: ABC with two angle bisectors: BD and BE.

Prove: mDBC = 0.

	Statements	Reasons
1.	BD and BE bisect ABC	Given
2.	ABC ~= DBC and ABE ~= EBC	Definition of angel bisector
3.	mABD = mDBC and mABE ~= mEBC	Definition of ~=
4.	mABD + mDBE + mEBC = mABC	Angle Addition Postulate
5.	mABD + mDBC = mABC and mABE + mEBC = mABC	Angle Addition Postulate
6.	2mABD = mABC and 2mEBC = mABC	Substitution (steps 3 and 5)
7.	mABD = mABC/2 and mEBC = mABC/2	Algebra
8.	mABC/2 + mDBE + mABC/2 = mABC	Substitution (steps 4 and 7)
9.	mABC + mDBE = mABC	Algebra
10.	mDBE = 0	Subtraction property of equality

4. The supplement of a right angle is a right angle.

A and B are supplementary angles, and A is a right angle.

Given: A and B are supplementary angles, and A is a right angle.

Prove: B is a right angle.

	Statements	Reasons
1.	A and B are supplementary angles, and A is a right angle	Given
2.	mA + mB = 180º	Definition of supplementary angles
3.	mA = 90º	Definition of right angle
4.	90º + mB = 180º	Substitution (steps 2 and 3)
5.	mB = 90º	Algebra
6.	B is a right angle	Definition of right angle

Proving Relationships Between Lines

m6 = 105º , m8 = 75º
Theorem 10.3: If two parallel lines are cut by a transversal, then the alternate exterior angles are congruent.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: 1 ~= 3.

	Statements	Reasons
1.	l m cut by a transversal t	Given
2.	1 and 2 are vertical angles	Definition of vertical angles
3.	2 and 3 are corresponding angles	Definition of corresponding angles
4.	2 ~= 3	Postulate 10.1
5.	1 ~= 2	Theorem 8.1
6.	1 ~= 3	Transitive property of 3.

3. Theorem 10.5: If two parallel lines are cut by a transversal, then the exterior angles on the same side of the transversal are supplementary angles.

l m cut by a transversal t.

Given: l m cut by a transversal t.

Prove: 1 and 3 are supplementary.

	Statement	Reasons
1.	l m cut by a transversal t	Given
2.	1 and 2 are supplementary angles, and m1 + m2 = 180º	Definition of supplementary angles
3.	2 and 3 are corresponding angles	Definition of corresponding angles
4.	2 ~= 3	Postulate 10.1
5.	m2 ~= m3	Definition of ~=
6.	m1 + m3 = 180º	Substitution (steps 2 and 5)
7.	1 and 3 are supplementary	Definition of supplementary

4. Theorem 10.9: If two lines are cut by a transversal so that the alternate exterior angles are congruent, then these lines are parallel.

Lines l and m are cut by a transversal t.

Given: Lines l and m are cut by a transversal t, with 1 ~= 3.

Prove: l m.

	Statement	Reasons
1.	Lines l and m are cut by a transversal t, with 1 ~= 3	Given
2.	1 and 2 are vertical angles	Definition of vertical angles
3.	1 ~= 2	Theorem 8.1
4.	2 ~= 3	Transitive property of ~=.
5.	2 and 3 are corresponding angles	Definition of corresponding angles
6.	l m	Theorem 10.7

5. Theorem 10.11: If two lines are cut by a transversal so that the exterior angles on the same side of the transversal are supplementary, then these lines are parallel.

Lines l and m are cut by a t transversal t.

Given: Lines l and m are cut by a transversal t, 1 and 3 are supplementary angles.

Prove: l m.

	Statement	Reasons
1.	Lines l and m are cut by a transversal t, and 1 are 3 supplementary angles	Given
2.	2 and 1 are supplementary angles	Definition of supplementary angles
3.	3 ~= 2	Example 2
4.	3 and 2 are corresponding angles	Definition of corresponding angles
5.	l m	Theorem 10.7

Two's Company. Three's a Triangle

An isosceles obtuse triangle
The acute angles of a right triangle are complementary.

ABC is a right triangle.

Given: ABC is a right triangle, and B is a right angle.

Prove: A and C are complementary angles.

	Statement	Reasons
1.	ABC is a right triangle, and B is a right angle	Given
2.	mB = 90º	Definition of right angle
3.	mA + mB + mC = 180º	Theorem 11.1
4.	mA + 90º + mC = 180º	Substitution (steps 2 and 3)
5.	mA + mC = 90º	Algebra
6.	A and C are complementary angles	Definition of complementary angles

3. Theorem 11.3: The measure of an exterior angle of a triangle equals the sum of the measures of the two nonadjacent interior angles.

ABC with exterior angle BCD.

	Statement	Reasons
1.	ABC with exterior angle BCD	Given
2.	DCA is a straight angle, and mDCA = 180º	Definition of straight angle
3.	mBCA + mBCD = mDCA	Angle Addition Postulate
4.	mBCA + mBCD = 180º	Substitution (steps 2 and 3)
5.	mBAC + mABC + mBCA = 180º	Theorem 11.1
6.	mBAC + mABC + mBCA = mBCA + mBCD	Substitution (steps 4 an 5)
7.	mBAC + mABC = mBCD	Subtraction property of equality

4. 12 units2

5. 30 units2

6. No, a triangle with these side lengths would violate the triangle inequality.

Congruent Triangles

1. Reflexive property: ABC ~= ABC.

Symmetric property: If ABC ~= DEF, then DEF ~= ABC.

Transitive property: If ABC ~= DEF and DEF ~= RST, then ABC ~= RST.

2. Proof: If ¯AC ~= ¯CD and ACB ~= DCB as shown in Figure 12.5, then ACB ~= DCB.

	Statement	Reasons
1.	¯AC ~= ¯CD and ACB ~= DCB	Given
2.	¯BC ~= ¯BC	Reflexive property of ~=
3.	ACB ~= DCB	SAS Postulate

3. If ¯CB ¯AD and ACB ~= DCB, as shown in Figure 12.8, then ACB ~= DCB.

	Statement	Reasons
1.	¯CB ¯AD and ACB ~= DCB	Given
2.	ABC and DBC are right angles	Definition of
3.	mABC = 90º and mDBC = 90º	Definition of right angles
4.	mABC = mDBC	Substitution (step 3)
5.	ABC ~= DBC	Definition of ~=
6.	¯BC ~= ¯BC	Reflexive property of ~=
7.	ACB ~= DCB	ASA Postulate

4. If ¯CB ¯AD and CAB ~= CDB, as shown in Figure 12.10, then ACB~= DCB.

	Statement	Reasons
1.	¯CB ¯AD and CAB ~= CDB	Given
2.	ABC and DBC are right angles	Definition of
3.	mABC = 90º and mDBC = 90º	Definition of right angles
4.	mABC = mDBC	Substitution (step 3)
5.	ABC ~= DBC	Definition of ~=
6.	¯BC ~= ¯BC	Reflexive property of ~=
7.	ACB ~= DCB	AAS Theorem

5. If ¯CB ¯AD and ¯AC ~= ¯CD, as shown in Figure 12.12, then ACB ~= DCB.

	Statement	Reasons
1.	¯CB ¯AD and ¯AC ~= ¯CD	Given
2.	ABC and DBC are right triangles	Definition of right triangle
3.	¯BC ~= ¯BC	Reflexive property of ~=
4.	ACB ~= DCB	HL Theorem for right triangles

6. If P ~= R and M is the midpoint of ¯PR, as shown in Figure 12.17, then N ~= Q.

	Statement	Reasons
1.	P ~= R and M is the midpoint of ¯PR	Given
2.	¯PM ~= ¯MR	Definition of midpoint
3.	NMP and RMQ are vertical angles	Definition of vertical angles
4.	NMP ~= RMQ	Theorem 8.1
5.	PMN ~= RMQ	ASA Postulate
6.	N ~= Q	CPOCTAC

Smiliar Triangles

x = 11
x = 12
40º and 140º
If A ~= D as shown in Figure 13.6, then BC/AB = CE/DE.

	Statement	Reasons
1.	A ~= D	Given
2.	BCA and DCE are vertical angles	Definition of vertical angles
3.	BCA ~= DCE	Theorem 8.1
4.	ACB ~ DCE	AA Similarity Theorem
5.	BC/AB = CE/DE	CSSTAP

5. 150 feet.

Opening Doors with Similar Triangles

If a line is parallel to one side of a triangle and passes through the midpoint of a second side, then it will pass through the midpoint of the third side.

¯DE ¯AC and D is the midpoint of ¯AB.

Given: ¯DE ¯AC and D is the midpoint of ¯AB.

Prove: E is the midpoint of ¯BC.

	Statement	Reasons
1.	¯DE ¯AC and D is the midpoint of ¯AB.	Given
2.	¯DE ¯AC and is cut by transversal AB	Definition of transversal
3.	BDE and BAC are corresponding angles	Definition of corresponding angles
4.	BDE ~= BAC	Postulate 10.1
5.	B ~= B	Reflexive property of ~=
6.	ABC ~ DBE	AA Similarity Theorem
7.	DB/AB = BE/BC	CSSTAP
8.	DB = AB/2	Theorem 9.1
9.	DB/AB = 1/2	Algebra
10.	1/2 = BE/BC	Substitution (steps 7 and 9)
11.	BC = 2BE	Algebra
12.	BE + EC = BC	Segment Addition Postulate
13.	BE + EC = 2BE	Substitution (steps 11 and 12)
14.	EC = BE	Algebra
15.	E is the midpoint of ¯BC	Definition of midpoint

2. AC = 43 , AB = 8 , RS = 16, RT = 83

3. AC = 42 , BC = 42

Putting Quadrilaterals in the Forefront

AD = 63, BC = 27, RS = 45
¯AX, ¯CZ, and ¯DY

Trapezoid ABCD with its XB CY four altitudes shown.

3. Theorem 15.5: In a kite, one pair of opposite angles is congruent.

Kite ABCD.

Given: Kite ABCD.

Prove: B ~= D.

	Statement	Reasons
1.	ABCD is a kite	Given
2.	¯AB ~= ¯AD and ¯BC ~= ¯DC	Definition of a kite
3.	¯AC ~= ¯AC	Reflexive property of ~=
4.	ABC ~= ADC	SSS Postulate
5.	B ~= D	CPOCTAC

4. Theorem 15.6: The diagonals of a kite are perpendicular, and the diagonal opposite the congruent angles bisects the other diagonal.

Kite ABCD.

Given: Kite ABCD

Prove: ¯BD ¯AC and ¯BM ~= ¯MD.

	Statement	Reasons
1.	ABCD is a kite	Given
2.	¯AB ~= ¯AD and ¯BC ~= ¯DC	Definition of a kite
3.	¯AC ~= ¯AC	Reflexive property of ~=
4.	ABC ~= ADC	SSS Postulate
5.	BAC ~= DAC	CPOCTAC
6.	¯AM ~= ¯AM	Reflexive property of ~=
7.	ABM ~= ADM	SAS Postulate
8.	¯BM ~= ¯MD	CPOCTAC
9.	BMA ~= DMA	CPOCTAC
10.	mBMA = mDMA	Definition of ~=
11.	MBD is a straight angle, and mBMD = 180º	Definition of straight angle
12.	mBMA + mDMA = mBMD	Angle Addition Postulate
13.	mBMA + mDMA = 180º	Substitution (steps 9 and 10)
14.	2mBMA = 180º	Substitution (steps 9 and 12)
15.	mBMA = 90º	Algebra
16.	BMA is a right angle	Definition of right angle
17.	¯BD ¯AC	Definition of

5. Theorem 15.9: Opposite angles of a parallelogram are congruent.

Parallelogram ABCD.

Given: Parallelogram ABCD.

Prove: ABC ~= ADC.

	Statement	Reasons
1.	Parallelogram ABCD has diagonal ¯AC.	Given
2.	ABC ~= CDA	Theorem 15.7
3.	ABC ~= ADC	CPOCTAC

6. 144 units2

7. 180 units2

8. Kite ABCD has area 48 units2.

Parallelogram ABCD has area 150 units2.

Rectangle ABCD has area 104 units2.

Rhombus ABCD has area 35/2 units2.

Anatomy of a Circle

Circumference: 20 feet, length of ˆRST = 155/18 feet
9 feet2
15 feet2
28º

The Unit Circle and Trigonometry

3/34 = 334/34
1/3 = 3/3
tangent ratio = 40/3, sine ratio = 40/7
tangent ratio = 5/56 = 556/56, cosine ratio = 56/9

Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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