CONGRUENT TRIANGLES

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1 ONGRUNT TRINGLS Two triangles are congruent if there is a sequence of rigid transformations that carry one onto the other. Two triangles are also congruent if they are similar figures with a ratio of similarity of 1, that is 1 1. One way to prove triangles congruent is to prove they are similar first, and then prove that the ratio of similarity is 1. In these lessons, students find short cuts that enable them to prove triangles congruent in fewer steps, by developing five triangle congruence conjectures. They are SSS, S, S, SS, and HL, illustrated below. SSS S S SS HL Note: S stands for side and stands for angle. HL is only used with right triangles. The H stands for hypotenuse and the L stands for leg. The pattern appears to be SS but this arrangement is NOT one of our conjectures, since it is only true for right triangles. See the Math Notes boxes in Lessons and ore onnections Geometry

2 xample 1 Use your triangle congruence conjectures to decide whether or not each pair of triangles must be congruent. ase each decision on the markings, not on appearances. Justify each answer. a. b. c. d. e. f. In part (a), the triangles are congruent by the SS conjecture. The triangles are also congruent in part (b), this time by the SSS conjecture. In part (c), the triangles are congruent by the S conjecture. art (d) shows a pair of triangles that are not necessarily congruent. The first triangle displays an S arrangement, while the second triangle displays an S arrangement. The triangles could still be congruent, but based on the markings, we cannot conclude that they definitely are congruent. The triangles in part (e) are right triangles and the markings fit the HL conjecture. Lastly, in part (f), the triangles are congruent by the S conjecture. arent Guide with xtra ractice 77

3 xample 2 Using the information given in the diagrams below, decide if any triangles are congruent, similar but not congruent, or not similar. If you claim the triangles are congruent or similar, create a flow chart justifying your answer. a. b. X Z V W Y In part (a), Δ Δ by the SS conjecture. Note: If you only see S, observe that is congruent to itself. The Reflexive roperty justifies stating that something is equal or congruent to itself. given = given = Δ Δ by SS Reflexive prop. In part (b), ΔWXV ~ ΔZYV by the ~ conjecture. The triangles are not necessarily congruent; they could be congruent, but since we only have information about angles, we cannot conclude anything else. Vertical angles = ΔWXV ~ ΔZYV ~ Lines, alt. int. angles = There is more than one way to justify the answer to part (b). There is another pair of alternate interior angles ( WXV and ZYV) that are equal that we could have used rather than the vertical angles, or we could have used them along with the vertical angles. 78 ore onnections Geometry

4 roblems riefly explain if each of the following pairs of triangles are congruent or not. If so, state the triangle congruence conjecture that supports your conclusion. F G J 3. K Q H I L V Y W M N O R S T T U X F 7. H I L N Q G J K M O 10. T U X 5 12 S 5 G H 3 4 R I W V R K 39º 112º J 112º 39º L Q F X W Y Z Use your triangle congruence conjectures to decide whether or not each pair of triangles must be congruent. ase your decision on the markings, not on appearances. Justify your answer Q S R arent Guide with xtra ractice 79

5 80 ore onnections Geometry Using the information given in each diagram below, decide if any triangles are congruent, similar but not congruent, or not similar. If you claim the triangles are congruent or similar, create a flowchart justifying your answer V W X Y Z T S K L R O N V I S L U N H T J K R I S

6 In each diagram below, are any triangles congruent? If so, prove it. (Note: Justify some using flowcharts and some by writing two-column proofs.) F omplete a proof for each problem below in the style of your choice. 32. Given: TR and MN bisect each other. 33. Given: bisects ; 1 2. rove: ΔNT ΔMR rove: Δ Δ N R T M Given:,, 35. Given: G SG, T TS rove: ΔF Δ rove: ΔTG ΔTSG F T G S 36. Given: O M, O bisects MO 37. Given:, rove: ΔMO ΔO rove: Δ Δ M O arent Guide with xtra ractice 81

7 38. Given: bisects, 39. Given: Q RS, R S rove: Δ Δ rove: ΔQR ΔQS 40. Given: S R, Q bisects SQR 41. Given: TU GY, KY HU, KT TG, rove: ΔSQ ΔRQ S HG TG. rove: K H K R Q S Q T Y U G R H 42. Given: MQ WL, MQ WL rove: ML WQ Q W M L onsider the diagram at right. 43. Is Δ Δ? rove it! 44. Is? rove it! 45. Is? rove it! 82 ore onnections Geometry

8 nswers 1. Δ ΔF by S. 2. ΔGIH ΔLJK by SS. 3. ΔNM ΔNO by SSS. 4. QS QS, so ΔQRS ΔQTS by HL. 5. The triangles are not necessarily congruent. 6. Δ ΔF by S or S. 7. GI GI, so ΔGHI ΔIJG by SSS. 8. lternate interior angles = used twice, so ΔKLN ΔNMK by S. 9. Vertical angles at 0, so ΔOQ ΔROS by SS. 10. Vertical angles and/or alternate interior angles =, so ΔTUX ΔVWX by S. 11. No, the length of each hypotenuse is different. 12. ythagorean Theorem, so ΔGH ΔIHG by SSS. 13. Sum of angles of triangle = 180º, but since the equal angles do not correspond, the triangles are not congruent. 14. F + F = F +, so Δ ΔF by SSS. 15. XZ XZ, so ΔWXZ ΔYXZ by S. 16. Δ Δ by S 17. ΔQS ΔRS by S, with S S by the Reflexive roperty. 18. ΔVXW ΔZXY by S, with VXW ZXY because vertical angles are. 19. ΔT ΔS by SSS, with by the Reflexive roperty. 20. ΔKL ΔL by HL, with L L by the Reflexive roperty. 21. Not necessarily congruent. 22. ΔV ~ ΔISV by SS ~ qual markings ΔV ~ ΔISV Vert. angles = SS~ 23. ΔLUN and ΔHT are not necessarily similar based on the markings. arent Guide with xtra ractice 83

9 24. ΔS ~ ΔSJ by ~. J given lt. int. angles = ΔS ~ ΔSJ ~ lt. int. angles = 25. ΔKRS ΔISR by HL given ΔKRS & ΔISR are right triangles. KR = IS given RS = RS ΔKRS ΔISR Refl. rop. HL 26. Yes 27. Yes Given Reflexive Δ Δ S right 's are Given vertical s are Given Δ Δ S 28. Yes 29. Yes Given Reflexive Δ Δ SS right 's are = Given Given Reflexive Δ Δ SSS 30. Not necessarily. 31. Yes ounterexample: F Given F Given Δ Δ F HL 32. N M and T R by definition of bisector. NT MR because vertical angles are equal. So, ΔNT ΔMR by SS. 33. by definition of angle bisector. by reflexive so Δ Δ by S. 34. since alternate interior angles of parallel lines congruent so ΔF Δ by S. 84 ore onnections Geometry

10 35. TG TG by reflexive so ΔTG ΔTSG by SSS. 36. MO O because perpendicular lines form right angles MO O by angle bisector and O O by reflexive. So, ΔMO ΔO by S. 37. and since parallel lines give congruent alternate interior angles. by reflexive so Δ Δ by S. 38. by definition of bisector. since vertical angles are congruent. So Δ Δ by S. 39. RQ SQ since perpendicular lines form congruent right angles. Q Q by reflexive so ΔQR ΔQS by S. 40. SQ RQ by angle bisector and Q Q by reflexive, so ΔSQ ΔRQ by S. 41. KYT HUG because parallel lines form congruent alternate exterior angles. TY + YU = YU + GU so TY GU by subtraction. T G since perpendicular lines form congruent right angles. So ΔKTY ΔHGU by S. Therefore, K H since triangles have congruent parts. 42. MQL WLQ since parallel lines form congruent alternate interior angles. QL QL by reflexive so ΔMQL ΔWLQ by SS so WQL MLQ since congruent triangles have congruent parts. So ML WQ since congruent alternate interior angles are formed by parallel lines. 43. Yes Reflexive 44. Not necessarily. 45. Not necessarily. Δ Δ SS arent Guide with xtra ractice 85

11 ONVRSS conditional statement is a sentence in the If then form. If all sides are equal in length, then a triangle is equilateral is an example of a conditional statement. We can abbreviate conditional statements by creating an arrow diagram. When the clause after the if in a conditional statement exchanges places with the clause after the then, the new statement is called the converse of the original. If the conditional statement is true, the converse is not necessarily true, and vice versa. See the Math Notes box in Lesson xample 1 Read each conditional statement below. Rewrite it as an arrow diagram, and state whether or not it is true. Then write the converse of the statement, and state whether or not the converse is true. a. If a triangle is equilateral, then it is equiangular. b. If x = 4, then x 2 = 16. c. If is a square, then is a parallelogram. The arrow diagram for part (a) is not much shorter than the original statement:δ is equilateral Δ is equiangular The converse is: If a triangle is equiangular, then it is equilateral. This statement and the original conditional statement are both true. In part (b), the conditional statement is true and the arrow diagram is: x = 4 x 2 = 16. The converse of this statement, If x 2 = 16, then x = 4, is not necessarily true because x could equal 4. In part (c), the arrow diagram is: is a square is a parallelogram. This statement is true, but the converse, If is a parallelogram, then is a square, is not necessarily true. It could be a parallelogram or a rectangle. 86 ore onnections Geometry

12 roblems Rewrite each conditional statement below as an arrow diagram and state whether or not it is true. Then write the converse of the statement and state whether or not the converse is true. 1. If an angle is a straight angle, then the angle measures If a triangle is a right triangle, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse. 3. If the measures of two angles of one triangle are equal to the measures of two angles of another triangle, then the measures of the third angles are also equal. 4. If one angle of a quadrilateral is a right angle, then the quadrilateral is a rectangle. 5. If two angles of a triangle have equal measures, then the two sides of the triangle opposite those angles have equal length. nswers 1. onditional: True 180 onverse: If an angle measures 180, then it is a straight angle. True. 2. onditional: True a c onverse: If the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse, then the triangle is a right triangle. True. 3. onditional: True b onverse: If the measures of one pair of corresponding angles of two triangles are equal, then the measures of the two other pairs of corresponding angles are also equal. False. arent Guide with xtra ractice 87

13 4. onditional: False is a rectangle onverse: If a quadrilateral is a rectangle, then one angle is a right angle. True, in fact, all four angles are right angles. 5. onditional: True onverse: If two sides of a triangle are equal in length, then the two angles opposite those sides are equal in measure. True. 88 ore onnections Geometry

14 LITIONS N ONNTIONS The remaining sections of hapter 6 are devoted to doing big problems. Students solve problems that involve many of the topics that they have studied so far, giving them the chance to connect the ideas and information as well as extend it to new situations. xample 1 To frame a doorway, strips of wood surround the opening creating a frame. If the doorway s dimensions are 30 inches by 80 inches and the strips of wood are inches wide, how much wood is needed to frame the doorway and how should it be cut? ssume that the strips will be assembled as shown in the figure at right and that they are sold in 8 lengths utting two pieces of wood 80 inches long, and one piece 30 inches long will not enable us to make a frame for the doorway. The inside edges of the strips of wood will have those measurements, but the outside dimensions of the wood are longer. The wood strips will meet at a 45 angle at the corners of the doorframe. Looking at the corner carefully (as shown at right) we see two triangles in the corner. The lengths and are both inches, since they are the width of the strips of wood. Since a triangle is isosceles, this means and are also inches in length. Therefore, we need two strips that are 82 1 inches long ( ), and one strip that is 35 inches long ( ), because each end must extend 45 the width of the vertical strip). Since the strips come in 8-foot lengths, we would need to buy three of them. Two will be cut at a 45 angle, with the outsides edge inches long and the inside edges 80 inches long. The third piece has two 45 angle cuts. The outside length is 35 inches while the inside length is 30 inches. arent Guide with xtra ractice 89

15 xample 2 friend offers to play a new game with you, using the spinner shown at right. Your friend says that you can choose to be player 1 or 2. On each turn, you will spin the spinner twice. If the letters are the same, player 1 gets a point. If the letters are different, then player 2 gets the point. Which player would you choose to be? Justify your answer. 120 To help us decide which player to be, we will create an area model to represent the probabilities. On the area model, the left edge represents the two outcomes from the first spin; the top edge represents the outcomes from the second spin. On the first spin, there are two possible outcomes, and, which are not equally likely. In fact, the probability of occurring is () = 2 3, while the probability of occurring is () = 1 3. This is true for the first spin and the second spin. We divide the area model according to these probabilities, and fill in the possible outcomes. layer 1 receives a point when the letters are the same for both spins. This outcome is represented by the shaded squares. layer 2 receives the point when the letters are different. y multiplying the dimensions of each region, the areas are expressed as ninths and we see that: (1 gets a point) = = 5 9 and (2 gets a point) = = 4 9 Since the probability that player 1 will win a point is greater than the probability that player 2 will, we should choose to be player ore onnections Geometry

16 roblems 1. On graph paper, plot the points (3, 4), (8, 1), (2, 9), and ( 3, 6) and connect them in order. Find all the measurements of this shape (side lengths, perimeter, area, and angle measures) and based on that information, decide the most specific name for this shape. Justify your answer. 2. The spinner at right is only partially completed. omplete the spinner based on these clues. a. There are three other single digit numbers on the spinner. ll four numbers on the spinner are equally likely results for one spin. No digit is repeated. 3 b. If the spinner is spun twice and the two outcomes are added, the largest possible sum is 16, while the smallest possible sum is 2. The most common sum is To go along with the snowflakes you are making for the winter dance decorating committee, you are going to make some Star olygons. Star olygon is formed by connecting equally spaced points on a circle in a specific order from a specified starting point. For instance, the circle at right has five equally spaced points. If we connect them in order, the shape is a regular pentagon (dashed sides). ut if we connect every other point, continuing until we reach the point that we started with, we get a star. a. What happens when 6 points are equally spaced around a circle? Under what conditions will you get a normal polygon, and when will you get a star polygon? start b. xplore other options. ome up with a rule that explains when a normal polygon is formed when connecting points, and when a star polygon is formed. onsider various numbers of points. start arent Guide with xtra ractice 91

17 nswers 1. The side lengths are: = = units, = = 136 = units. erimeter = units. The slope of = the slope of = 5 3, slope of = slope of = 5 3. Since the slopes are negative reciprocals, we know that the segments are perpendicular, so all four angles are 90, so the figure is a rectangle. rea = 68 square units. 2. The spinner is divided into four equal pieces, with the numbers 3, 1, 6, and 8. start 3. (a) onnecting consecutive points forms a hexagon. onnecting every other point forms an equilateral triangle. onnecting every third point forms several line segments (diameters), but no star. (b) Trying the same things with 7 points around a circle produces more interesting results. onnecting each point in order forms a heptagon (7-gon). onnecting every second point or every fifth point produces the first star polygon at right. onnecting every third point or every fourth point produces the second star polygon at right. To generalize, if the number of points around the circle is n, and we connect to the r th point, the polygon is a star polygon if n and r have no common factors. Note that whenever r = 1 or r = n 1, the result is always a polygon. very second point or every fifth point very third point or every fourth point start start 92 ore onnections Geometry