Triangles

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Triangles Sections Definition of a Triangle Triangle Inequality Triangle Interior Angle Sum Interior, Exterior, Remote Interior Angles Corresponding Parts Included Parts Triangle Congruence Theorems Perpendicular Bisector Theorem Angle Bisector Theorem Circumcenter Circumscribed / Inscribed Incenter Median Centroid Theorem Midsegment Theorem Hinge Theorem Pythagorean Theorem More on Triangle Side Length Relationships Special Right Triangles Orthocenter Triangle Similarity Triangle Proportionality Right Triangle Similarity Theorem Triangle Trigonometry Problems Problem Solutions

Definition of a Triangle A triangle is a closed two-dimensional (flat) shape composed of three line segments, A corner of a triangle is called a vertex. A triangle has three vertices. A vertex is a point.

There are types of triangles, which are; Acute: each angle is acute. Obtuse: one angle is obtuse. Right: one angle is 90o, a right angle. When drawing a right triangle, the 90o angle is represented by a tiny box. Isosceles: the base angles are congruent and the base angle opposite sides are congruent. Scalene: none of the sides have the same length. Equilateral: each angle has a measure of 60o and the sides are congruent.

Triangle Inequality For a triangle that has side lengths a, b and c, the inequality is a + b > c. Any three line segments can form a triangle when positioned properly:

(See problem 26)

Triangle Interior Angle Sum

The triangle interior angle sum can be shown by drawing an auxiliary line parallel to the base, extending the sides, then using congruent angles on parallel lines and vertical angles. The angles on the auxiliary parallel line will be congruent to the corresponding interior angles and will sum to 180o. (In the Proofs Study Guide, see problem 1 there)

Interior, Exterior, Remote Interior Angles

From angle d, the Remote Interior angles are angle b and angle c. Because a triangle’s interior angles sum to 180o, and d + a = 180o, 180o = 180o a+b+c=a+d b+c=d (See problem 1) Remote Interior Angle Sum: d = b + c (See problem 1)

Corresponding Parts For two triangles; a) The bases are in corresponding positions. b) The left sides are in corresponding positions. c) The right sides are in corresponding positions. d) The base angles are in corresponding positions. e) The top angles are in corresponding positions. These parts, from different triangles, but having the same triangle-part position, are called corresponding parts. (See problems 2–12)

Included Parts A triangle part that is between two other triangle parts is an included part.

BAC is included between and BAC. (See problems 4, 5, 11)

and

.

is includedbetween

ABC

Triangle Congruence Theorems Congruence theorems are rules that are used to confirm that two or more triangles are congruent. a) Side - Side - Side (SSS): For two triangles that have all corresponding sides congruent, the two triangles are congruent. The triangles fit on top of each other. b) Side - Angle - Side (SAS): For two triangles that have two corresponding sets of congruent parts that are composed of two consecutive sides and an included angle, the triangles are congruent. The corresponding included angles causes the third corresponding sides that closes the triangles to be congruent. c) Angle - Side - Angle (ASA): For two triangles that have two corresponding sets of congruent parts that are composed of two consecutive angles and an included side, the triangles are congruent. The corresponding congruent angles cause the corresponding second and third sides to be congruent. d) Angle - Angle - Side (AAS): For two triangles that have two corresponding sets of congruent parts that are composed of two angles and a nonincluded side, the triangles are congruent. The corresponding congruent nonincluded sides close the triangles. The sides opposite the corresponding congruent angles have to be congruent for the nonincluded sides to close the triangles. e) Hypotenuse - Leg (HL): For two right triangles that have two corresponding sets of congruent parts that are composed of a hypotenuse and a leg, the triangles are congruent. The Pythagorean Theorem calculates the length of a hypotenuse. For the known length of the congruent hypoteni and the known length

of the congruent legs, the other corresponding legs have to be the same length for the Pythagorean Theorem equation to hold true for the congruent right triangles. (See problems 2-7, 9, 11, 41)

Perpendicular Bisector Theorem A line, line segment, or ray, that is perpendicular to a second line segment and bisects the second line is a perpendicular bisector. The line segment showing the height of an isosceles triangle bisects the base, so this line segment is a perpendicular bisector. This also shows the end points of the bisected line segment are equidistant to a common point on the perpendicular bisector. (See problems 13, 14, 39)

Angle Bisector Theorem A line, line segment, or ray, that bisects an angle is an angle bisector. The shortest distance from an angle-interior point on the angle bisector and the sides of the bisected angle is shown by the line segments perpendicular to the sides of the angle that extend from the angle-interior point on the angle bisector. This results in two congruent right triangles sharing a common hypotenuse, the two non-base corresponding angles sharing a common vertex, and the two right base angles being in corresponding positions having different vertices. (See problem 13)

Circumcenter A circumcenter is a point associated with a triangle that is equidistant from the vertices of the triangle. The circumcenter can be outside of the triangle. (See problems 15, 18, 19)

Circumscribed / Inscribed A circle that is outside of a triangle such that the all three vertices of the triangle touches the circle is called a circumscribed circle. A circle that is inside of a triangle and touches all three sides is called an inscribed circle. (See problems 15, 18)

Incenter The center of an inscribed circle is called the incenter. The incenter is the point of currency for the segments that are both perpendicular to the sides of a triangle and bisect the angles of the triangle. (See problem 15)

Median A median is a segment that extends from a vertex of a triangle to the midpoint of the side opposite the vertex. A triangle has three medians. (See problem 21)

Centroid Theorem The point on all three medians of a triangle that is the distance from any vertex is called the centroid. A triangle has only one centroid. (See problem 22 below, the is proof shown in the Proofs Study Guide, see problem 11 there)

Midsegment Theorem A segment that a) Extends from one midpoint of the side of a triangle to the opposite midpoint on the opposite side of the triangle, b) Is parallel to the base of the triangle, c) Is the length of the base of the triangle is called the midsegment of the triangle. (See problems 23, 24, 55, proof shown in the Proofs Study Guide, there see problem 21)

Hinge Theorem Let two triangles have corresponding sets of two consecutive sides and an included angle, and the corresponding sides are congruent while the corresponding angles are not. The corresponding angles have sides opposite them. These sides opposite the corresponding angles are in corresponding positions. The corresponding side opposite the smaller angle is shorter than the other corresponding side opposite the larger corresponding angle in the other triangle. Prove BH > BC. Given:

1) (top-left side) moves closer to (bottom side) as A decreases in size, because an angle changes measure when the one of the angle sides changes position. 2) (top-left side) moves further from (bottom side) as A increases in size, because an angle changes measure when the one of the angle sides change position. 3) BD < BE < BF, because of statements 1) and 2). 4) DC < EC < FC, because of statements 1) and 2). 5) is a hypotenuse, because BDC, BEC, and BFC are right triangles. 6) Let BC = smallest BC, BH = longest BC, BG = second longest BC. 7) BC < BG < BH, because of statements 3), 4), and the Pythagorean Theorem. (See problems 27-29)

Pythagorean Theorem For a right triangle having hypotenuse of length c, base of length b, and height of length a, c2 = a2 + b2 This can be solved for c to find the length of the hypotenuse,

(See problems 16, 19, 20 below, for the proof see the Proofs Study Guide, problem 22 there)

More on Triangle Side Length Relationships For a triangle having side lengths c, a, and b; c is the longest side. If c2 > a2 + b2, the triangle is an obtuse triangle. If c2 < a2 + b2, the triangle is an acute triangle. These triangle side relationships can be realized using a right triangle and the Pythagorean Theorem as the reference equation. If the Pythagorean Theorem equation is not satisfied, the triangle is not a right triangle. If a triangle is not a right triangle, the angle is either an acute or obtuse triangle. (See problems 29, 30)

Special Right Triangles

3 - 4 - 5 right triangle

(See problems 16, 17) Verification of the 45o - 45o - 90o right triangle: 1) Let a square have a side length of 1. 2) A line segment from a corner across the square’s interior to another corner is a diagonal. 3) The diagonal bisects the angles of the corners. 4) Two right triangles are then formed, each triangle having angles of 45o, 45o, 90o, and a height of length 1, and a base of length 1. 5) The diagonal is a hypotenuse. 6) The Pythagorean Theorem shows the hypotenuse to have a length of . Verification of the 30o - 60o - 90o right triangle: 1) Let an equilateral triangle have a side length of 2. 2) The perpendicular bisector of a side bisects the angle opposite the midpoint of the side that is bisected, so each half of the side has a length of 1. 3) The two halves of the bisected angle are 30o. 4) Two right triangles are then formed, each having angles of 30o, 60o, 90o, a base of length 1, and a

hypotenuse of length 2. 5) The perpendicular bisector is the height of the right triangles formed. 6) The Pythagorean Theorem show the height to have a length of . Verification of the 3 - 4 - 5 right triangle: 1) The Triangle Inequality shows sides of lengths 3, 4, and 5 can form a triangle, 3 + 4 > 5. 2) The Pythagorean Theorem shows a triangle having side lengths of 3, 4, and 5 can form a right triangle. 3) The longest side of length 5 has to be the hypotenuse. 4) 52 = 32 + 42 and 25 = 9 + 16.

Orthocenter The orthocenter of a triangle is the intersection of the heights of the triangle. From each side of a triangle, a height can be drawn. To find orthocenter P of triangle ABC: 1) Find the slopes of the non-vertical and non-horizontal sides. 2) Find the perpendicular slopes on the sides from step 1). 3) Find the equations of the lines having perpendicular slopes and opposite vertices. 4) For triangles having a vertical or horizontal side, let this side’s equation be a coordinate of P. 5) Test possible intersections of lines having the perpendicular slopes. (See problems 64-66)

Triangle Similarity There are three triangle similarity postulates; Angle – Angle (AA): If two triangles have congruent corresponding pairs of angles, then the triangles are similar. Side – Side – Side (SSS): If all sides of two triangles are proportional, then the triangles are similar. Side – Angle – Side (SAS): For ABC, (See problems 23, 24, 31-35, 37, 45, 46, 51-53)

Triangle Proportionality

(See problems 23, 31, 36, 47-50)

Right Triangle Similarity Theorem

Geometric Mean: For positive numbers a and b, From the right triangle similarity theorem,

(See problems 40-44)

Triangle Trigonometry Trigonometry is the study of angles for various measurements. Triangles are used. A right triangle shows trigonometric ratios. The initial ratios are called sine, cosine, and tangent . For an angle x, the ratios are sin(x), cos(x), and tan(x), as shown on most scientific calculators. These ratio names are set equal to ratios composed of the triangle side lengths.

Observe that x only equals the angle measure, never any of the ratios. For a right triangle, to use the trigonometric ratios to find the angle measures, inverse functions are needed. For the same right triangle above:

For an acute or obtuse triangle, angles and side lengths can be found using the Law of Sines or Law of Cosines:

(See problems 56-63)

Problems 1. Find x. The angle units are degrees:

2. If the triangles are congruent, tell why. The angle units are degrees.

3. Explain if the triangles are congruent or not.

4. Explain if the triangles are congruent or not.

5. Explain if the triangles are congruent or not. The angle units are degrees.

6. Explain if the triangles are congruent or not.

7. Explain why the triangles are congruent.

8. Why is there not enough details to tell if the triangles are congruent?

9. Explain for which case the two triangles can be proven to be congruent.

10. Why are the triangles not congruent? The angle units are degrees.

11.

DBC is isosceles. Show

ADB

CDE.

12. ABC is equilateral. Show degrees.

13. For

ADB

ABC, what is true about

14. For equilateral

ACE. The angle units are

?

ABC, what is true about

,

, and

?

15. What point on the interior of a triangle is central to a triangle?

16. Find x.

17. Find c. The units for the angles are degrees.

18.

XZW is equilateral. Find the distance XY.

19. A triangle has side lengths of 4. Find the length of the line segments from the vertices to the circumcenter G. 20. If a circle is inscribed inside of a 3 – 4 – 5 right triangle, what is the diameter of the circle? 21. A triangle has vertices (1, 1), (4, 6), (8, 3). Determine the length of the median from (1, 1) to the opposite side. 22. A triangle has vertices (0, 0), (5, 8), (9, 0). Determine the length of the segments from the vertices to the centroid P.

25. For ABC and DEF, Compare BC and EF.

26. Solve for x:

27. Solve for x:

and

correspond, and m(

A) > m(

D).

28. Prove Theorem.

, which is the final result of the Triangle Angle Bisector

29. For a triangle that has side lengths of 10, 12, and 15, determine if the triangle is obtuse, acute or right. 30. For a triangle that has side lengths of 9, 14, and 16, determine if the triangle is obtuse, acute or right. 31. Prove: , which is the final result of the Two-Transversal Proportionality Corollary.

32. Prove: For two similar triangles, the ratio of their corresponding angle bisectors is the same as the ratio of their corresponding sides. Given:

33. Show the Side-Angle-Side (SAS) Similarity Postulate is true.

34. Show the Angle-Angle (AA) Similarity Postulate is true. The same illustration from problem #35 will be referred to.

35. Show the Side-Side-Side (SSS) Similarity Postulate is true. The same illustration from problem #35 will be referred to. 36. Find x and y.

37. Show

ABC ~

ABD ~

ADC.

38. Find x.

39. Show the Perpendicular Bisector Theorem is true.

40. Find AD.

41. Find AB.

42. Find AC.

43. Find BD.

44. Find DC.

45. Find x.

46. Explain if the triangles are similar or not.

47. Find AD and DC.

48. Find x and AC.

49. Find DC and AB.

50. Find x.

51. For the similar triangles, find x.

52. Explain if the triangles are similar or not.

53. Explain if the triangles are similar or not.

54. Find DE.

55. FBDE is a parallelogram, and AE.

56. Find sin(B).

57. Find cos(C).

is the midsegment. Show that FE =

58. Find tan(B).

59. Find cos(x) and x.

60. Find a, b, m(

61. Find c.

A ).

62. Find x.

63. Using the height of a triangle, show that the Law of Sines is true. 64. Find the orthocenter of the triangle having vertices (-4, 2), (-2, 6), (2, 2).

65. Find the orthocenter of the triangle having vertices (-2, 1), (3, -3), (4, 4).

66. Find the orthocenter of the triangle having vertices ( 2, -2), (4, 6), (8, -2).

Problem Solutions 1. Find x. The angle units are degrees:

Using the Remote Interior Angle Sum, 130 = x + 65 130 - 65 = x 65 = x 2. If the triangles are congruent, tell why. The angle units are degrees.

The triangles are congruent because the congruent corresponding sides having known length are included between two corresponding pairs of congruent angles, which is the ASA congruence postulate. 3. Explain if the triangles are congruent or not.

1) The non-base corresponding sides of the isosceles triangles are congruent. 2) . 3) The triangles are congruent because all corresponding sides are congruent, which is the SSS congruence postulate. 4. Explain if the triangles are congruent or not.

1) and are nonincluded, corresponding and congruent. 2) BDA and A are a consecutive pair. 3) BDC and C are a consecutive pair. 4) BDA BDC, A C. 5) The consecutive angle pairs of both triangles are corresponding. 6) The triangles are congruent by the AAS postulate. 5. Explain if the triangles are congruent or not. The angle units are degrees.

1) For the first triangle, the length of the included side between the known angels is not known. 2) The length of the included side between the known angles of the second triangle is known. 3) The first triangle can be repositioned so the known angles of both sides are clearly in corresponding positions.

4) The corresponding included sides between the corresponding pairs of angles are not congruent. 5) The triangles are not congruent because the ASA postulate is not satisfied. 6. Explain if the triangles are congruent or not.

1) The opposite sides of a rectangle are congruent. 2) The diagonal is congruent to itself. 3) Then ABC DCA by either the SSS or HL postulate 4) The opposite angles of a rectangle are congruent. 5) Then ABC DCA by the SAS postulate. 6) The diagonal divides the angels it intersects into congruent opposite interior angles because the rectangle’s diagonal is a transversal on the rectangles opposite parallel sides. . 7) Then ABC DCA is congruence by the ASA or AAS postulate. 7. Explain why the triangles are congruent.

1) An angle measures the amount of opening, and the marked angles are congruent. 2) The marked angles open the correct same amount, and the vertices of the marked angles are far enough away from the opposite sides, for the angles to have the opposite side perfectly fit. 3) The sides that are opposite the marked angles are congruent.

4) The corresponding pair of unmarked sides of the triangles are the same length because of statement 2). 5) The triangles are congruent by the SSS postulate because of statements 3) and 4). 8. Why is there not enough details to tell if the triangles are congruent?

1) The lengths of the sides of an angle do not affect the measure of the angle. 2) The lengths of the sides of one of the triangles can be different than the side lengths of the other triangle.

9. Explain for which case the two triangles can be proven to be congruent.

1) For case 1, the congruent parts are corresponding because the second triangle can be flipped at D over the first triangle, positioning D over C and F over A. 2) For case 2, the second triangle can be flipped at D over the first triangle, positioning D over C and F over A, but the congruent sides will never be in corresponding positions. 3) Congruent parts have to be in corresponding positions. Only for case 1 can the triangles be proven to be congruent using the AAS postulate.

10. Why are the triangles not congruent? The angle units are degrees.

1) The angles and sides are in their corresponding positions. 2) For DEF to remain closed and the angles remain the same, DE > AB and DF > AC because EF > BC. 3) The sides of DEF have to be longer than the sides of ABC, causing the triangles to not be congruent. 11.

DBC is isosceles. Show

ADB

CDE.

1) A and E are corresponding and congruent. 2) B and C are corresponding and congruent. 3) and are corresponding and congruent, and each side is not included between the corresponding congruent angles. 4) The triangles are congruent by the AAS postulate. 12. ABC is equilateral. Show degrees.

ADB

ACE. The angle units are

1) m( DBA) + m( CBA) = m( DBA) + 60 = 180. 2) m( ECA) + m( BCA) = m( ECA) + 60 = 180. 3) m( DBA) = m( ECA) = 120. 4) DBA ECA, because they are both angles and they both measure o the same 120 . 5) DBA and ECA are in corresponding positions. 6) and are in corresponding positions and are congruent. 7) and are in corresponding positions and are congruent. 8) ADB ACE by the SAS postulate. 13. For

ABC, what is true about

?

1) B is equidistant from A and from C. 2) m( ) is the shortest distance from B to D considering that AB > BD and BC > BD. 3) satisfies the Perpendicular Bisector criteria, so and bisects ABC. 14. For equilateral

ABC, what is true about

,

, and

?

1) Theorem. 2) Theorem. 3) Theorem.

and

bisects

BAC from the Perpendicular Bisector

and

bisects

BCA from the Perpendicular Bisector

and

bisects

ABC from the Perpendicular Bisector

15. What point on the interior of a triangle is central to a triangle?

1) Case 1 shows the circumcenter of a triangle, found by a circumscribed circle. 2) Case 2 shows the incenter of a triangle, found by an inscribed circle. 3) Case 3 shows the circumcenter and the incenter for an equilateral triangle are in the same position. , , and each are the height of the triangle and each are medians of the Triangle. From the definition of the radius of a circle and the Centroid Theorem,

of

each of these line segments result in the same length, so

,

, and

are each of the height of the triangle and are each part of the line segment that shows the height of the triangle. 4) The two central points of a triangle are the circumcenter and the incenter and only for an equilateral triangle are the two points in the same position. 16. Find x.

1) For the right triangle the Pythagorean Theorem can be used to find the length of the hypotenuse. 2) Let c represent the hypotenuse. 3) c = = = o o o 4) The 45 - 45 - 90 triangle is the only right triangle that has side lengths of 1, 1, and . 5) The small box represents 90o. 6) x = 45o 17. Find c. The units for the angles are degrees.

1) The only right triangle that has angle measures of 30o, 60o, and 90o is the 30o - 60o - 90o right triangle.

2) The hypotenuse of a 30o - 60o - 90o right triangle is 2 3) c is in the hypotenuse position, so c = 2. 18.

XZW is equilateral. Find the distance XY.

1) UYV is a right triangle because radii are along the sides of quarters of the circles. 2) VY = 10 because it is a radius. 3) passes through the circumcenter of equilateral XZW, so bisects ZXW. 4) || . 5) m( ZXW) = m( ZXW) = 30o. 6) UYV is a 30o - 60o 90o right triangle. 7) Because is opposite the 30oangle, VY = 10 = 1(10), and UYV has a scale factor of 10. 8) UV = circumcenter of

, UY = 10 + n = 2(10), n = distance from the XZW to Y.

9) UY = + = from step 8), 10 + n = 2(10) n = 20 - 10 = 10. 10) XY = 10 + 10 + 10 = 30

=

= 20, so n = 10, also

19. A triangle has side lengths of 4. Find the length of the line segments from the vertices to the circumcenter G.

20. If a circle is inscribed inside of a 3 – 4 – 5 right triangle, what is the diameter of the circle?

21. A triangle has vertices (1, 1), (4, 6), (8, 3). Determine the length of the median from (1, 1) to the opposite side.

22. A triangle has vertices (0, 0), (5, 8), (9, 0). Determine the length of the segments from the vertices to the centroid P.

This reasoning can be applied to line segments that are parallel to the base of ACE and have endpoints that are not midpoints of the sides of ACE.

25. For ABC and DEF, Compare BC and EF.

and

correspond, and m(

A) > m(

D).

26. Solve for x:

27. Solve for x:

28. Prove Theorem.

, which is the final result of the Triangle Angle Bisector

4) BAD DAC, because A for BAC is given as bisected 5) XAY DAC, because the angles are vertical angles. 6) BAD YAZ, because the angles are vertical angles. 7) XAY YAZ, because of steps 3) and 4). 8) BXA YAX, because the angels are alternate interior angles on parallel rays. 9) XBA YAZ, because the angles are corresponding angles on parallel rays. 10) AXB ABX, because of steps 6) and 7). 11) AXB is isosceles, because of step 8).

29. For a triangle that has side lengths of 10, 12, and 15, determine if the triangle is obtuse, acute or right. 1) Let c = 15, a = 10, b = 144 2) 152 = 225, 102 = 100, 122 = 144. 3) 225 ≠ 100 + 144, so the triangle is not a right triangle 4) 225 > 244, so the triangle is an obtuse triangle. 30. For a triangle that has side lengths of 9, 14, and 16, determine if the triangle is obtuse, acute or right.

1) Let c = 16, a = 9, b = 14 2) 92 = 81, 142 = 196, 162 = 256 3) 256 ≠ 81 + 196, so the triangle is not a right triangle 4) 256 < 277, so the triangle is an acute triangle

31. Prove: , which is the final result of the Two-Transversal Proportionality Corollary.

32. Prove: For two similar triangles, the ratio of their corresponding angle bisectors is the same as the ratio of their corresponding sides. Given:

33. Show the Side-Angle-Side (SAS) Similarity Postulate is true.

1)Because the sides of ACE can be scaled by the same scale factor g to the size of the sides of BCD, the corresponding sides of the triangles are proportional, and the angle measures remain the same. a = gd, b = ge, c = gf. 2) The single scale factor g allows for proportionality. 3) The detail of the corresponding congruent angles having to be included between the sides that the proportion is composed of ensures that these sides are in corresponding positions, else the corresponding angles could become non-congruent upon closing the triangles. 34. Show the Angle-Angle (AA) Similarity Postulate is true. The same illustration from problem #35 will be referred to.

1) If two angles of ACE are congruent to two angles of BCD, the third angle of both triangles will be congruent. 2) For congruent corresponding angels of both triangles, the sides of the corresponding angles will be parallel upon necessary alignment, because the direction of the sides of an angle determine the size of an angle. 3) Corresponding angles of the triangles have corresponding sides. 4) The sides of ACE can be scaled by the same scale factor to the size of sides of BCD, because the corresponding sides are parallel. 35. Show the Side-Side-Side (SSS) Similarity Postulate is true. The same illustration from problem #35 will be referred to. 1) If the corresponding sides of ACE and BCD are proportional, then ACE can be scaled to the size of BCD. 2) If the sides of a triangle are all scaled to a different size by the same scale factor, the angles will remain the same size because angle side length does not affect angle size. 36. Find x and y.

Using the Triangle Proportionality properties, the needed proportions can be written then solved. After solving the first proportion for x, substitute the found x-value into the second proportion then solve for y. 37. Show

ABC ~

ABD ~

ADC.

1) ABC is a 45o- 45o- 90o triangle. 2) ADC is a 45o- 45o- 90o triangle. 3) ABC is a 45o- 45o- 90o triangle. 4) All three triangles are similar to each other by the Angle-Angle Similarity Postulate. 38. Find x.

39. Show the Perpendicular Bisector Theorem is true.

1)

ADX

CDX, because of the Side-Angle-Side Postulate.

3) AD = DC because of step 2). 40. Find AD.

41. Find AB.

42. Find AC.

43. Find BD.

44. Find DC.

45. Find x.

46. Explain if the triangles are similar or not.

47. Find AD and DC.

48. Find x and AC.

49. Find DC and AB.

50. Find x.

51. For the similar triangles, find x.

52. Explain if the triangles are similar or not.

53. Explain if the triangles are similar or not.

54. Find DE.

55. FBDE is a parallelogram, and AE.

56. Find sin(B).

is the midsegment. Show that FE =

57. Find cos(C).

58. Find tan(B).

59. Find cos(x) and x.

60. Find a, b, m(

61. Find c.

A ).

62. Find x.

63. Using the height of a triangle, show that the Law of Sines is true.

64. Find the orthocenter of the triangle having vertices (-4, 2), (-2, 6), (2, 2).

Steps 1) and 2):

Step 3): From to B, the perpendicular line has the equation x = -2.

From to A, the perpendicular line has the equation y – 2 = 1(x – [-4]), y=x+4+2 y=x+6 From

to C, the perpendicular line has the equation y – 2 =

(x –

2), y=

x+1+2

y=

x+3

Steps 4) and 5): From step 3), x = -2. Find orthocenter P = (-2, y). On the perpendicular line passing through A, P = (-2, -2 + 6) = (-2, 4). On the perpendicular line passing through C, P = (-2,

(-2) + 3) = (-2,

4). P = (-2, 4) is on the perpendicular line passing through B. Orthocenter = P = (-2, 4). 65. Find the orthocenter of the triangle having vertices (-2, 1), (3, -3), (4, 4).

Steps 1) and 2):

Step 3): From

to B, the perpendicular line has the equation y – 4 =

y–4=

x–5

y=

x–5+4

y=

x-1

From [-2]),

to A, the perpendicular line has the equation y – 1 =

y–1= y=

(x – 4),

(x –

x– x–

+

y= x+ From to C, the perpendicular line has the equation y – (-3) = -2(x – 3), y + 3 = -2x + 6 y = -2x + 3 Step 4): For the intersection of the perpendicular lines from to B, x+ = x–1 -4x + 20 = 35x – 28 48 = 39x

to A and from

=x x= For the intersection of the perpendicular lines from to C,

to B and from

-2x + 3 = x – 1 -8x + 12 = 5x – 4 16 = 13x =x For the intersection of the perpendicular lines from to C,

to A and from

x + = -2x + 3 -x + 5 = -14x + 21 13x = 16 x= For the perpendicular line from

to A, y =

·

+

For the perpendicular line from

to B, y =

·

–1=

For the perpendicular line from

to C, y = -2·

+3=

=

Step 5): Orthocenter P = (

,

)

66. Find the orthocenter of the triangle having vertices ( 2, -2), (4, 6), (8, -2).

Steps 1) – 2):

Step 3): From

to A, the perpendicular line has the equation y – (-2) =

y+2=

x–1

(x –

2),

y=

x–3

From

to C, the perpendicular line has the equation y – (-2) =

(x

– 8), y+2= y=

x+2 x

Step 4): For the intersection of the perpendicular lines from to C, x=4

to B and from

P = (4, [4]) = (4, -1) For the intersection of the perpendicular lines from to B, x=4 P = (4, [4] – 3) = (4, 2 - 3) = (4, -1) Step 5): P = (4, -1)

to A and from