Given: and are Given: P Given: Given: and are Given: and Given: and are

19. and

20. and

21. and

22. and

In Exercises 23 to 25, prove the indicated relationship. c = 8 a = 7 b = 4 a = 5 c = 10 b = 6 c = 17 a = 15

35. Given: Regular pentagon ABCDE with diagonals

and Prove: HINT: First prove .

36. In the figure with regular

pentagon ABCDE, do and trisect ? HINT: . m ∠ABE = m∠AEB ∠ABC BD BE 䉭 ABE ⬵ 䉭CBD BE ⬵ BD BD BE

33. As a car moves along

the roadway in a mountain pass, it passes through a horizontal run of 750 feet and through a vertical rise of 45 feet. To the nearest foot, how far does the car move along the roadway?

34. Because of construction along the road from A to B,

Alinna drives 5 miles from A to C and then 12 miles from C to B. How much farther did Alinna travel by using the alternative route from A to B?

32. In the support system of the bridge shown, and

. Find: a b c BS m ∠ABD m ∠RST m ∠ABC = 28° AC = 6 ft

30. Given: and

Prove: HINT: First show that .

31. In the roof truss shown, and

. Find: a AH b c m ∠ADB m ∠BAD m ∠HAF = 37° AB = 8 䉭 ACE ⬵ 䉭ADB 䉭BDC ⬵ 䉭ECD AB ⬵ AE CE ⬜ DE DB ⬜ BC

26. Given: and are

rt. Prove: HINT: Show .

27. Given: and

Prove: HINT: Show .

28. Given: and

Prove: HINT: Show .

29. Given: bisects

Prove: HINT: First show that . 䉭 RSW ⬵ 䉭RUW 䉭 TRU ⬵ 䉭VRS RS ⬵ RU ∠SRU RW 䉭 MQP ⬵ 䉭PNM MQ ⬵ NP MQ 7 NP MN 7 QP 䉭 NMP ⬵ 䉭QPM MQ 7 NP MN ⬵ QP ∠1 ⬵ ∠2 䉭 MQP ⬵ 䉭NPQ MP ⬵ NQ MQ ⬵ NP ∠s ∠NPQ ∠MQP

23. Given: and

Prove: bisects

24. Given: bisects

Prove: E is the midpoint of

25. Given: E

is the midpoint of Prove: In Exercises 26 to 28, draw the triangles that are to be shown congruent separately. DE ⬜ FG DF ⬵ DG FG FG ∠F ⬵ ∠G ∠FDG DE ∠FDG DE FE ⬵ EG DF ⬵ DG C A a c b B F D G E Exercises 23–25 Q M P N 1 3 2 4 Exercises 26–28 V T R U S W Exercise 29 G E F D B H C A D C B S V R T A Run Rise C B A A B C D E Exercises 35, 36 D C B E A In an isosceles triangle, the two sides of equal length are legs, and the third side is the base. See Figure 3.22. The point at which the two legs meet is the vertex of the trian- gle, and the angle formed by the legs and opposite the base is the vertex angle. The two remaining angles are base angles. If in Figure 3.23, then is isosceles with legs and , base , vertex C, vertex angle C, and base angles at A and B. With , we see that the base of this isosceles triangle is not neces- sarily the “bottom” side. AB AC ⬵ BC AB BC AC 䉭ABC AC ⬵ BC Isosceles Triangle Vertex, Legs, and Base of an Isosceles Triangle Base Angles Vertex Angle Angle Bisector Median Altitude Perpendicular Bisector Auxiliary Line Determined, Overdetermined, Undetermined Equilateral and Equiangular Triangles Perimeter Isosceles Triangles 3.3 KEY CONCEPTS Leg Leg Vertex Vertex Angle Base Base Angles Figure 3.22 A a 1 2 D B ⬔ 1 ⫽ ⬔2, so AD is the angle-bisector of ⬔BAC in ⌬ABC ~ C Figure 3.23 M is the midpoint of BC, so AM is the median from A to BC B C A M b AE BC, so AE is the altitude of ⌬ABC from vertex A to BC A E B C c M is the midpoint of BC and FM BC, so FM is the perpendicular bisector of side BC in ⌬ABC M F B C A d Consider in Figure 3.23 once again. Each angle of a triangle has a unique angle bisector, and this may be indicated by a ray or segment from the vertex of the bisected angle. Just as an angle bisector begins at the vertex of an angle, the median also joins a ver- tex to the midpoint of the opposite side. Generally, the median from a vertex of a triangle is not the same as the angle bisector from that vertex. An altitude is a line segment drawn from a vertex to the opposite side such that it is perpendicular to the opposite side. Finally, the perpendicular bisector of a side of a triangle is shown as a line in Figure 3.23d. A seg- ment or ray could also perpendicularly bisect a side of the triangle. In Figure 3.24, is the bisector of ; is the altitude from A to ; M is the midpoint of ; is the median from A to and is the perpendicular bisector of . An altitude can actually lie in the exterior of a triangle. In Figure 3.25 on page 146, which shows obtuse triangle , the altitude from R must be drawn to an exten- sion of side . Later we will use the length h of the altitude and the length b of side in the following formula for the area of a triangle: Any angle bisector and any median necessarily lie in the interior of the triangle. A = 1 2 bh ST RH ST 䉭RST BC Í FM BC ; AM BC BC AE ∠BAC AD 䉭ABC B E C D M F A Figure 3.24