1. Chapter 1 Congruent Triangles

2. In this case, we write and we say that the various congruent angles and segments "correspond" to each other. DEFINITION. Two triangles ABC and DEF are said to be congruent if the following six conditions all hold :

4. SAS THEOREM (POSTULATE) If and, then.

6. If and, then.

7. Proof of ASA Assume that in and and. Consider now and. If we knew that they were congruent, then we would be done: The two triangles would be congruent by SAS.

8. So we assume that they are not equal and is longer. Since is longer there is an interior point G on such that. But now and satisfy: and So, by SAS,.

9. Hence, But, by hypothesis But this is impossible: has to be smaller than because G is an interior point of Hence, This is a contradiction. our assumption that is not congruent tomust beFalse. Therefore, and the proof is concluded.. Comparing these two equations, we see that

10. Let be any triangle 1. If then 2. If then Isosceles Triangle Theorem

11. Proof. (1): In this case we have these three Congruences,, and So Hence, as corresponding part of congruent triangles. by ASA (!).

12. Proof. (2): Now use the congruences,, And. To conclude that by SAS. So.

13. SSS THEOREM If and, then.

14. Assume in and GWe now construct a point such that that, and By ASA,.. To prove the theorem we will prove that

15. Construct. by assumption, Since and as corresponding parts of congruent triangles,, by transitivity. Hence is an isosceles triangle. So,. we conclude that

17. By a similar line of reasoning isosceles triangle and is an Now consider the following relations: Together,these imply that This is impossible. Hence,G must coincide with A, and by SAS.

19. PROBLEM. To bisect an angle. Solution. First, use your compass to construct B and C such that. Then draw an arc with center B,and arc with center C, both with the same radius. Let D be their point of intersection. Then is the angle bisector.

21. Since and We may conclude that by SSS. Hence,.

22. PROBLEM Given angle and ray, construct an, angle that will be congruent to.

24. Proof Since We have by SSS. Hence.

25. PROBLEM Given a line and a point P not on. construct a line that passes through P and is perpendicular to.

27. Proof Consider the two triangles and. They must be congruent by SSS. From this we conclude that Now, let C be the point of intersection of andand considerand These two triangles haveas a common side, they haveand they have

28. congruent angles at P, as noted previously. HenceSo, But, These two equations imply that and,as claimed...

29. AB A B A B PP Q Q Perpendicular Bisector of Segment

31. The Vertical Angel Theorem Let the lines and meet at point P, as shown in the figure. Then and B C A D P

32. Proof Clearly, and Hence, So The case of is similar.

33. THE EXTERIOR ANGLE THEOREM In e x tend to a point D on, forming the exterior angle. Then is greater than each of and, the remote interior angles. B A E F CD

34. Proof The first part of the proof will consist of constructing the picture in figure. First, let E be the midpoint of B to E and ; then connect extend to a point F in the interior of such thatNow consider the. triangles and.

35. By construction, they have two of their respective sides congruent. Moreover, by the vertical angle theorem. Hence. This implies that. But is less than. So as claimed The case ofis similar..

36. be a triangle in which Let is longer than.. Then. or, the greater angle is opposite the greater side. B CD A

37. Proof We may find a point D on such that. Since is isosceles,. By the exterior angle theorem and, clearly, Comparing these three statements, the result follows.

38. THEOREM Let be a triangle in which Then. or, the greater side is opposite the greater angle. BC A

39. If the theorem was not true, then either or.. If,then the triangle would be isosceles and we would get the contradiction. If,then.,which is also a contradiction.

40. THEOREM In any triangle, the sum of the lengths of any two sides is greater than the length of the third side. D A B C

41. be any triangle. We will show that. to a point D such that. So. now, since is isosceles, so. Therefore in is a side opposite a larger angle Extend the side Let. The theorem now follows.