||
is read “line is parallel to line .” In diagrams, arrows are used to indicate that lines are parallel (see Fig. 4-1).Parallel lines are straight lines which lie in the same plane and do not intersect however far they are extended. The symbol for parallel is ||; thus, || is read “line is parallel to line .” In diagrams, arrows are used to indicate that lines are parallel (see Fig. 4-1).
Fig. 4-1
A transversal of two or more lines is a line that cuts across these lines. Thus, 
is a transversal of 
and , in Fig. 4-2.
Fig. 4-2
The interior angles formed by two lines cut by a transversal are the angles between the two lines, while the exterior angles are those outside the lines. Thus, of the eight angles formed by and cut by in Fig. 4-2, the interior angles are ∠1, ∠2, ∠3, and ∠4; the exterior angles are ∠5, ∠6, ∠7, and ∠8.
Corresponding angles of two lines cut by a transversal are angles on the same side of the transversal and on the same side of the lines. Thus, ∠1 and ∠2 in Fig. 4-3 are corresponding angles of and cut by trans-versal . Note that in this case the two angles are both to the right of the transversal and both below the lines.
Fig. 4-3
When two parallel lines are cut by a transversal, the sides of two corresponding angles form a capital F in varying positions, as shown in Fig. 4-4.
Fig. 4-4
Alternate interior angles of two lines cut by a transversal are nonadjacent angles between the two lines and on opposite sides of the transversal. Thus, ∠1 and ∠2 in Fig. 4-5 are alternate interior angles of and cut by . When parallel lines are cut by a transversal, the sides of two alternate interior angles form a capital Z or N in varying positions, as shown in Fig. 4-6.
Fig. 4-5
Fig. 4-6
When parallel lines are cut by a transversal, interior angles on the same side of the transversal can be readily located by noting the capital U formed by their sides (Fig. 4-7).
Fig. 4-7
PRINCIPLE 1: Through a given point not on a given line, one and only one line can be drawn parallel to a given line. (Parallel-Line Postulate)
Thus, either l1 or l2 but not both may be parallel to l3 in Fig. 4-8.
Fig. 4-8
Proving that Lines are Parallel
PRINCIPLE 2: Two lines are parallel if a pair of corresponding angles are congruent.
Thus, l1 || l2 if ∠ a
∠ b in Fig. 4-9.
Fig. 4-9
PRINCIPLE 3: Two lines are parallel if a pair of alternate interior angles are congruent.
Thus, l1 || l2 if ∠ c ∠ d in Fig. 4-10.
Fig. 4-10
PRINCIPLE 4: Two lines are parallel if a pair of interior angles on the same side of a transversal are supplementary.
Thus, l1 || l 2 if ∠e and ∠f are supplementary in Fig. 4-11.
Fig. 4-11
PRINCIPLE 5: Lines are parallel if they are perpendicular to the same line. (Perpendiculars to the same line are parallel.)
Thus, l1 || l 2 if l 1 and l 2 are each perpendicular to l 3 in Fig. 4-12.
Fig. 4-12
PRINCIPLE 6: Lines are parallel if they are parallel to the same line. (Parallels to the same line are parallel.)
Thus, l1 || l2 if l1 and l2 are each parallel to l3 in Fig. 4-13.
Fig. 4-13
Properties of Parallel Lines
PRINCIPLE 7: If two lines are parallel, each pair of corresponding angles are congruent. (Corresponding angles of parallel lines are congruent)
Thus, if l1 || l2, then ∠A = ∠b in Fig. 4-14.
Fig. 4-14
PRINCIPLE 8: If two lines are parallel, each pair of alternate interior angles are congruent. (Alternate interior angles of parallel lines are congruent.)
Thus, if l1 || l2, then ∠c ∠d in Fig. 4-15.
Fig. 4-15
PRINCIPLE 9: If two lines are parallel, each pair of interior angles on the same side of the transversal are supplementary.
Thus, if l1|| l2, ∠e and ∠f are supplementary in Fig. 4-16.
Fig. 4-16
PRINCIPLE 10: If lines are parallel, a line perpendicular to one of them is perpendicular to the others also.
Thus, if l1 || l 2 and l 3 
l 1, then l3 l2 in Fig. 4-17.
Fig. 4-17
PRINCIPLE 11: If lines are parallel, a line parallel to one of them is parallel to the others also.
Thus, if l1 || l2 and l3 || l1 then l3 || l2 in Fig. 4-18.
Fig. 4-18
PRINCIPLE 12: If the sides of two angles are respectively parallel to each other, the angles are either congruent or supplementary.
Thus, if l1 || l3 and l2 || l4 in Fig. 4-19, then ∠A ∠b and ∠A and ∠c are supplementary.
Fig. 4-19
4.1 Numerical applications of parallel lines
In each part of Fig. 4-20, find the measure x and the measure y of the indicated angles.
Fig. 4-20
Solutions
(a) x= 130° (Principle 8). y = 180°- 130°= 50° (Principle 9).
(b) x = 80° (Principle 8). y = 70° (Principle 7).
(c) x = 75° (Principle 7). y = 180°– 75°= 105° (Principle 9).
(d) x = 65° (Principle 7). Since m∠B = m∠A, m∠B = 65°. Hence, y = 65° (Principle 8).
(e) x = 30° (Principle 8). y= 180°– (30°+ 70°) = 80° (Principle 9).
(f) x = 180°– 110°= 70° (Principle 9). y = 110° (Principle 12).
4.2 Applying parallel line principles and their converses
The following short proofs refer to Fig. 4-21. In each, the first statement is given. State the parallel-line principle needed as the reason for each of the remaining statements.
Fig. 4-21
Solutions
(a) 2: Principle 3; 3: Principle 7.
(b) 2: Principle 2; 3: Principle 9.
(c) 2: Principle 4; 3: Principle 8.
(d) 2: Principle 5; 3: Principle 10.
4.3 Algebraic applications of parallel lines
In each part of Fig. 4-22, find x and y. Provide the reason for each equation obtained from the diagram.
Fig. 4-22
Solutions
4.4 Proving a parallel-line problem
PROOF:
4.5 Proving a parallel-line problem stated in words
Prove that if the diagonals of a quadrilateral bisect each other, the opposite sides are parallel.
PROOF:
The distance between two geometric figures is the straight line segment which is the shortest segment between the figures.
1. The distance between two points, such as P and Q in Fig. 4-23(a), is the line segment 
between them.
Fig. 4-23
2. The distance between a point and a line, such as P and in (b), is the line segment PQ, the perpendicular from the point to the line.
3. The distance between two parallels, such as and in (c), is the segment , a perpendicular between the two parallels.
4. The distance between a point and a circle, such as P and circle O in (d), is , the segment of 
between the point and the center of the circle.
5. The distance between two concentric circles, such as two circles whose center is O, is , the segment of the larger radius that lies between the two circles, as shown in (e).
PRINCIPLE 1: If a point is on the perpendicular bisector of a line segment, then it is equidistant from the ends of the line segment.
Thus if P is on , the bisector of 
in Fig. 4-24, then 
≐ 
.
Fig. 4-24
PRINCIPLE 2: If a point is equidistant from the ends of a line segment, then it is on the perpendicular bisector of the line segment. (Principle 2 is the converse of Principle 1.)
Thus if ≐ in Fig. 4-24, then P is on , the bisector of .
PRINCIPLE 3: If a point is on the bisector of an angle, then it is equidistant from the sides of the angle.
Thus if P is on , the bisector of ∠A in Fig. 4-25, then ≐ 
, where PQ and PR are the distances of P from the sides of the angle.
Fig. 4-25
PRINCIPLE 4: If a point is equidistant from the sides of an angle, then it is on the bisector of the angle. (Principle 4 is the converse of Principle 3.)
Thus if PQ = PR, where PQ and PR are the distances of P from the sides of ∠A in Fig. 4-25, then P is on , the bisector of ∠A.
PRINCIPLE 5: Two points each equidistant from the ends of a line segment determine the perpendicular bisector of the line segment. (The line joining the vertices of two isosceles triangles having a common base is the perpendicular bisector of the base.)
Thus if ≐ and 
≐ 
in Fig. 4-26, then P and Q determine , the bisector of .
Fig. 4-26
PRINCIPLE 6: The perpendicular bisectors of the sides of a triangle meet in a point which is equidistant from the vertices of the triangle.
Thus if P is the intersection of the bisectors of the sides of 
ABC in Fig. 4-27, then ≐ ≐ 
. P is the center of the circumscribed circle and is called the circumcenter of ABC.
Fig. 4-27
PRINCIPLE 7: The bisectors of the angles of a triangle meet in a point which is equidistant from the sides of the triangle.
Thus if Q is the intersection of the bisectors of the angles of ABC in Fig. 4-28, then 
≐ 
≐ 
, where these are the distances from Q to the sides of ABC. Q is the center of the inscribed circle and is called the incenter of ABC.
Fig. 4-28
4.6 Finding distances
In each of the following, find the distance and indicate the kind of distance involved. In Fig. 4-29(a), find the distance (a) from P to A; (b) from P to ; (c) from A to 
(d) from to . In Fig. 4-29(b), find the distance (e) from P to inner circle O; (f) from P to outer circle O; (g) between the concentric circles.
Fig. 4-29
Solutions
(a) PA = 7, distance between two points
(b) PG = 4, distance from a point to a line
(c) AE = 10, distance from a point to a line
(d) FG = 6, distance between two parallel lines
(e) PQ =12 – 3 = 9, distance from a point to a circle
(f) PR = 12 – 5 = 7, distance from a point to a circle
(g) QR = 5 – 3 = 2, distance between two concentric circles
4.7 Locating a point satisfying given conditions
In Fig. 4-30.
(a) Locate P, a point on and equidistant from A and C.
(b) Locate Q, a point on and equidistant from 
and 
.
(c) Locate R, the center of the circle circumscribed about ABC.
(d) Locate S, the center of the circle inscribed in ABC.
Fig. 4-30
See Fig. 4-31.
Fig. 4-31
4.8 Applying principles 2 and 4
For each ABC in Fig. 4-32, describe P, Q, and R as equidistant points, and locate each on a bisector.
Fig. 4-32
Solutions
(a) Since P is equidistant from A and B, it is on the bisector of . Since Q is equidistant from B and C, it is on the bisector of BC. Since R is equidistant from A, B, and C, it is on the bisectors of , , and .
(b) Since P is equidistant from and , it is on the bisector of ∠B. Since Q is equidistant from 
and , it is on the bisector of ∠C. Since R is equidistant from , , and , it is on the bisectors of ∠A, ∠B, and ∠C.
4.9 Applying principles 1, 3, 6, and 7
For each ABC in Fig. 4-33, describe P, Q, and R as equidistant points. Also, describe R as the center of a circle.
Fig. 4-33
(a) Since P is on the bisector of ∠A, it is equidistant from and . Since Q is on the bisectors of ∠B, it is equidistant from and . Since R is on the bisectors of ∠A and ∠B, it is equidistant from , , and . R is the incenter of ABC, that is, the center of its inscribed circle.
(b) Since P is on the bisector of , it is equidistant from A and B. Since Q is on the bisector of , it is equidistant from A and C. Since R is on the bisectors of and , it is equidistant from A, B, and C. R is the circumcenter of ABC, that is, the center of its circumscribed circle.
4.10 Applying principles 1, 3, 6, and 7
In each part of Fig. 4-34, find two points equidistant from the ends of a line segment, and find the perpendicular bisector determined by the two points.
Fig. 4-34
Solutions
(a) B and D are equidistant from A and C; hence, 
is the bisector of .
(b) A and D are equidistant from B and C; hence, 
is the bisector of .
(c) B and D are equidistant from A and C; hence, 
is the bisector of . A and C are equidistant from B and D; hence, is the bisector of .
The angles of any triangle may be torn off, as in Fig. 4-35(a), and then fitted together as shown in (b). The three angles will form a straight angle.
We can prove that the sum of the measures of the angles of a triangle equals 180° by drawing a line through one vertex of the triangle parallel to the side opposite the vertex. In Fig. 4-35(c), 
is drawn through B parallel to AC. Note that the measure of the straight angle at B equals the sum of the measures of the angles of ABC; that is, a° + b° + c° = 180°. Each pair of congruent angles is a pair of alternate interior angles of parallel lines.
Fig. 4-35
An exterior angle of a polygon is formed whenever one of its sides is extended through a vertex. If each of the sides of a polygon is extended, as shown in Fig. 4-36, an exterior angle will be formed at each vertex. Each of these exterior angles is the supplement of its adjacent interior angle.
Fig. 4-36
Thus, in the case of pentagon ABCDE, there will be five exterior angles, one at each vertex. Note that each exterior angle is the supplement of an adjacent interior angle. For example, m∠a + m∠a' = 180°.
PRINCIPLE 1: The sum of the measures of the angles of a triangle equals the measure of a straight angle or 180°.
Thus in ABC of Fig. 4-37, m∠A + m∠B + m∠C = 180°.
Fig. 4-37
PRINCIPLE 2: If two angles of one triangle are congruent respectively to two angles of another triangle, the remaining angles are congruent.
Thus in ABC and A′B′C in Fig. 4-38, if LA ≡ LA′ and LB ≡ LB′, then LC ≡ LC′.
Fig. 4-38
PRINCIPLE 3: The sum of the measures of the angles of a quadrilateral equals 360°.
Thus in quadrilateral ABCD (Fig. 4-39), m∠A + m∠B + m∠C + mLD = 360°.
Fig. 4-39
PRINCIPLE 4: The measure of each exterior angle of a triangle equals the sum of the measures of its two nonadjacent interior angles.
Thus in ABC in Fig. 4-40, mLECB = m∠A + m∠B.
Fig. 4-40
PRINCIPLE 5: The sum of the measures of the exterior angles of a triangle equals 360°.
Thus in ABC in Fig. 4-41, mZa’ + m∠b’+ m∠C′= 360°.
Fig. 4-41
PRINCIPLE 6: The measure of each angle of an equilateral triangle equals 60°.
Thus if ABC in Fig. 4-42 is equilateral, then m∠A = 60°, m∠B = 60°, and m∠C = 60°.
Fig. 4-42
PRINCIPLE 7: The acute angles of a right triangle are complementary.
Thus in rt. ABC in Fig. 4-43, if m∠C = 90°, then m∠A + m∠B = 90°.
Fig. 4-43
PRINCIPLE 8: The measure of each acute angle of an isosceles right triangle equals 45°.
Thus in isos. rt. ABC in Fig. 4-44, if m∠C = 90°, then m∠A = 45° and m∠B = 45°.
Fig. 4-44
PRINCIPLE 9: A triangle can have no more than one right angle.
Thus in rt. ABC in Fig. 4-43, if m∠C = 90°, then ∠A and ∠B cannot be rt.
PRINCIPLE 10: A triangle can have no more than one obtuse angle.
Thus in obtuse ABC in Fig. 4-45, if ∠C is obtuse, then ∠A and ∠B cannot be obtuse angles.
Fig. 4-45
PRINCIPLE 11: Two angles are supplementary if their sides are respectively perpendicular to each other.
Thus if l1 l3 and l2 l4 in Fig. 4-46, then ∠A ∠b, and a and ∠c are supplementary.
Fig. 4-46
4.11 Numerical applications of angle-measure-sum principles
In each part of Fig. 4-47, find x and y.
Fig. 4-47
Solutions
Check: The sum of the measures of the angles of quad. ABCD should equal 360°.
4.12 Applying angle-measure-sum principles to isosceles and equilateral triangles
Find x and y in each part of Fig. 4-48.
Fig. 4-48
Solutions
4.13 Applying ratios to angle-measure sums
Find the measure of each angle
(a) Of a triangle if its angle measures are in the ratio of 3:4:5 [Fig. 3-49(a)]
(b) Of a quadrilateral if its angle measures are in the ratio of 3:4:5:6 [(b)]
(c) Of a right triangle if the ratio of the measures of its acute angles is 2:3 [(c)]
Fig. 4-49
(a) Let 3x, 4x, and 5x represent the measures of the angles. Then 12x = 180 by Principle 1, so that x = 15. Now 3x = 45, 4x = 60, and 5x = 75. Ans. 45°, 60°, 75°
(b) Let 3x, 4x, 5x, and 6x represent the measures of the angles. Then 18x = 360 by Principle 3, so that x = 20. Now 3x = 60, 4x = 80, and so forth. Ans. 60°, 80°, 100°, 120°
(c) Let 2x and 3x represent the measures of the acute angles. Then 5x = 90 by Principle 7 so that x = 18. Now 2x = 36 and 3x = 54. Ans. 36°, 54°, 90°
4.14 Using algebra to prove angle-measure-sum problems
(a) Prove that if the measure of one angle of a triangle equals the sum of the measures of the other two, then the triangle is a right triangle.
(b) Prove that if the opposite angles of a quadrilateral are congruent, then its opposite sides are parallel.
Solutions
A polygon is a closed plane figure bounded by straight line segments as sides. An n-gon is a polygon of n sides. Thus, a polygon of 20 sides is a 20-gon.
Fig. 4-50
A regular polygon is an equilateral and equiangular polygon. Thus, a regular pentagon is a polygon having 5 congruent angles and 5 congruent sides (Fig. 4-50). A square is a regular polygon of 4 sides.
Names of Polygons According to the Number of Sides
By drawing diagonals from any vertex to each of the other vertices, as in Fig. 4-51, a polygon of 7 sides is divisible into 5 triangles. Note that each triangle has one side of the polygon, except the first and last triangles which have two such sides.
In general, this process will divide a polygon of n sides into n – 2 triangles; that is, the number of such triangles is always two less than the number of sides of the polygon.
The sum of the measures of the interior angles of the polygon equals the sum of the measures of the interior angles of the triangles. Hence:
Sum of measures of interior angles of a polygon of n sides = (n–2)180°
Fig. 4-51
The exterior angles of a polygon can be reproduced together, so that they have the same vertex. To do this, draw lines parallel to the sides of the polygon from a point, as shown in Fig. 4-52. If this is done, it can be seen that regardless of the number of sides, the sum of the measures of the exterior angle equals 360°. Hence:
Sum of measures of exterior angles of a polygon of n sides = 360°
Fig. 4-52
For any polygon
PRINCIPLE 1: If S is the sum of the measures of the interior angles of a polygon of n sides, then
The sum of the measures of the interior angles of a polygon of 10 sides (decagon) equals 1440°, since S = 8(180) = 1440.
PRINCIPLE 2: The sum of the measures of the exterior angles of any polygon equals 360°.
Thus, the sum of the measures of the exterior angles of a polygon of 23 sides equals 360°.
For a regular polygon
PRINCIPLE 3: If a regular polygon of n sides (Fig. 4-53) has an interior angle of measure i and an exterior angle of measure e (in degrees), then
Fig. 4-53
Thus for a regular polygon of 20 sides,
4.15 Applying angle-measure formulas to a polygon
(a) Find the sum of the measures of the interior angles of a polygon of 9 sides (express your answer in straight angles and in degrees).
(b) Find the number of sides a polygon has if the sum of the measures of the interior angles is 3600°.
(c) Is it possible to have a polygon the sum of whose angle measures is 1890°?
Solutions
(a) S (in straight angles) = n – 2 = 9 – 2 = 7 straight angles; m ∠ S = (n – 2)180 = 7(180) = 1260°.
(b) S (in degrees) = (n – 2)180. Then 3600 = (n – 2)180, from which n = 22.
(c) Since 1890 = (n- 2)180, then n = 12 
. A polygon cannot have 12 sides.
4.16 Applying angle-measure formulas to a regular polygon
(a) Find each exterior angle measure of a regular polygon having 9 sides.
(b) Find each interior angle measure of a regular polygon having 9 sides.
(c) Find the number of sides a regular polygon has if each exterior angle measure is 5°.
(d) Find the number of sides a regular polygon has if each interior angle measure is 165°.
Solutions
(a) Since n = 9, m∠e =
= 40. Ans. 40°
(b) Since n = 9, m∠i =
= 140. Ans. 140°
Another method: Since i + e = 180, i = 180–e = 180–40 = 140.
(c) Substituting e = 5 in 
, we have 
. Then 5n = 360, so n = 72. Ans. 72 sides
(d) Substituting i = 165 in i = e = 180, we have 165 + e = 180 or e = 15.
Then, using 
with e = 15, we have 
or n = 24. Ans. 24 sides
4.17 Applying algebra to the angle-measure sums of a polygon
Find each interior angle measure of a quadrilateral (a) if its interior angles are represented by x + 10, 2x + 20, 3x - 50, and 2x - 20; (b) if its exterior angles are in the ratio 2:3:4:6.
(a) Since the sum of the measures of the interior 
is 360°, we add
Then x + 10 = 60; 2x + 20 = 120; 3x–50 = 100; 2x–20 = 80. Ans. 60°, 120°, 100°, 80°
(b) Let the exterior angles be represented respectively by 2x, 3x, 4x, and 6x. Then 2x + 3x + 4x + 6x = 360. Solving gives us 15x = 360 and x = 24. Hence, the exterior angles measure 48°, 72°, 96°, and 144°. The interior angles are their supplements. Ans. 132°, 108°, 84°, 36°
Three methods of proving triangles congruent have already been introduced here. These are
1. Side-Angle-Side (SAS)
2. Angle-Side-Angle (ASA)
3. Side-Side-Side (SSS)
Two additional methods of proving that triangles are congruent are
4. Side-Angle-Angle (SAA)
5. Hypotenuse-Leg (hy-leg)
PRINCIPLE 1: (SAA) If two angles and a side opposite one of them of one triangle are congruent to the corresponding parts of another, the triangles are congruent.
Thus if ∠A = ∠A’, ∠B ≐ ∠B’, and ≐ 
in Fig. 4-54, then ABC = AA’B’C’.
Fig. 4-54
PRINCIPLE 2: (hy-leg) If the hypotenuse and a leg of one right triangle are congruent to the corresponding parts of another right triangle, the triangles are congruent.
Thus if hy ≐ by 
and leg ≐ leg
in Fig. 4-55, then rt. ABC ≐ rt. A′B′C.
Fig. 4-55
A proof of this principle is given in Chapter 16.
4.18 Selecting congruent triangles using hy-leg or SAA
In (a) Fig. 4-56 and (b) Fig. 4-57, select congruent triangles and state the reason for the congruency.
Fig. 4-56
Fig. 4-57
Solutions
(a) I AII by hy-leg. In III, 4 is not a hypotenuse.
(b) I AIII by SAA. In II, 7 is opposite 60° instead of 35°. In IV, 7 is included between 60° and 35°.
4.19 Determining the reason for congruency of triangles
In each part of Fig. 4-58, I can be proved congruent to II. Make a diagram showing the congruent parts of both triangles and state the reason for the congruency.
Fig. 4-58
(a) See Fig. 4-59(a). I AII by hy-leg.
(b) See Fig. 4-59(b). I AII by SAA.
Fig. 4-59
4.20 Proving a congruency problem
PROOF:
4.21 Proving a congruency problem stated in words
Prove that in an isosceles triangle, altitudes to the congruent sides are congruent.
4.1. In each part of Fig. 4-60, find x and y.
(4.1)
Fig. 4-60
4.2. In each part of Fig. 4-61, find x and y.
(4.2)
Fig. 4-61
4.3. If two parallel lines are cut by a transversal, find
(4.3)
(a) Two alternate interior angles represented by 3x and 5x–70
(b) Two corresponding angles represented by 2x + 10 and 4x–50
(c) Two interior angles on the same side of the transversal represented by 2x and 3x
4.4. Provide the proofs requested in Fig. 4-62
(4.4)
Fig. 4-62
4.5. Provide the proofs requested in Fig. 4-63
(4.4)
Fig. 4-63
4.6. Prove each of the following:
(4.5)
(a) If the opposite sides of a quadrilateral are parallel, then they are also congruent.
(b) If and 
bisect each other at E, then || .
(c) In quadrilateral ABCD, let || . If the diagonals and intersect at E and 
|| 
, then ≐ 
.
(d) and are parallel lines cut by a transversal at E and F. If 
and 
bisect a pair of corresponding angles, then || .
(e) If a line through vertex B of ABC is parallel to and bisects the angle formed by extending through B, then ABC is isosceles.
4.7. In Fig. 4-64, find the distance from (a) A to B; (b) E to (c) A to (d) 
to .
(4.6)
Fig. 4-64
Fig. 4-65
4.8. In Fig. 4-65, find the distance (a) from P to the outer circle; (b) from P to the inner circle; (c) between the concentric circles; (d) from P to O.
(4.6)
4.9. In Fig. 4-66
(4.6)
(a) Locate P, a point on equidistant from B and C. Then locate Q, a point on , equidistant from and .
(b) Locate R, a point equidistant from A, B, and C. Then locate S, a point equidistant from B, C, and D.
(c) Locate T, a point equidistant from , and . Then locate U, a point equidistant from , and .
Fig. 4-66
Fig. 4-67
4.10. In each part of Fig. 4-67, describe P, Q, and R as equidistant points and locate them on a bisector.
4.11. In each part of Fig. 4-68, describe P, Q, and R as equidistant points.
(4.9)
Fig. 4-68
4.12. Find x and y in each part of Fig. 4-69.
(4.11)
Fig. 4-69
4.13. Find x and y in each part of Fig. 4-70.
(4.12)
Fig. 4-70
4.14. Find the measure of each angle
(4.13)
(a) Of a triangle if its angle measures are in the ratio 1:3:6
(b) Of a right triangle if its acute angle measures are in the ratio 4:5
(c) Of an isosceles triangle if the ratio of the measures of its base angle to a vertex angle is 1:3
(d) Of a quadrilateral if its angle measures are in the ratio 1:2:3:4
(e) Of a triangle, one of whose angles measures 55° and whose other two angle measures are in the ratio 2:3
(f) Of a triangle if the ratio of the measures of its exterior angles is 2:3:4
4.15. Prove each of the following:
(4.14)
(a) In quadrilateral ABCD if ∠A= ∠d and ∠b = ∠C, then || .
(b) Two parallel lines are cut by a transversal. Prove that the bisectors of two interior angles on the same side of the transversal are perpendicular to each other.
4.16. Show that a triangle is
(4.14)
(a) Equilateral if its angles are represented by x + 15, 3x–75, and 2x–30
(b) Isosceles if its angles are represented by x + 15, 3x–35, and 4x
(c) A right triangle if its angle measures are in the ratio 2:3:5
(d) An obtuse triangle if one angle measures 64° and the larger of the other two measures 10° less than five times the measure of the smaller
4.17. (a) Find the sum of the measures of the interior angles (in straight angles) of a polygon of 9 sides; of 32 sides.
(4.15)
(b) Find the sum of the measures of the interior angles (in degrees) of a polygon of 11 sides; of 32 sides; of 1002 sides.
(c) Find the number of sides a polygon has if the sum of the measures of the interior angles is 28 straight angles; 20 right angles; 4500°; 36,000°.
4.18. (a) Find the measure of each exterior angle of a regular polygon having 18 sides; 20 sides; 40 sides.
(4.16)
(b) Find the measure of each interior angle of a regular polygon having 18 sides; 20 sides; 40 sides.
(c) Find the number of sides a regular polygon has if each exterior angle measures 120°; 40°; 18°; 2°.
(d) Find the number of sides a regular polygon has if each interior angle measures 60°, 150°; 170°, 175°; 179°.
4.19. (a) Find each interior angle of a quadrilateral if its interior angles are represented by x–5, x + 20, 2x– 45, and 2x– 30.
(4.17)
(b) Find the measure of each interior angle of a quadrilateral if the measures of its exterior angles are in the ratio of 1:2:3:3.
4.20. In each part of Fig. 4-71, select congruent triangles and state the reason for the congruency.
(4.18)
4.21. In each part of Fig. 4-72, two triangles can be proved congruent. Make a diagram showing the congruent parts of both triangles, and state the reason for the congruency. (4.19)
(4.19)
Fig. 4-71
Fig. 4-72
4.22. Provide the proofs requested in Fig. 4-73.
(4.20)
Fig. 4-73
4.23. Prove each of the following:
(4.21)
(a) If the perpendiculars to two sides of a triangle from the midpoint of the third side are congruent, then the triangle is isosceles.
(b) Perpendiculars from a point in the bisector of an angle to the sides of the angle are congruent.
(c) If the altitudes to two sides of a triangle are congruent, then the triangle is isosceles.
(d) Two right triangles are congruent if the hypotenuse and an acute angle of one are congruent to the corresponding parts of the other.
