Establishing a further relationship, like bisects

The following theorem can be applied only when a triangle is a right triangle contains a right angle. Proof of the theorem is delayed until Section 5.4. Figure 3.20 Finally, with point A as center, mark off a length equal to that of as shown. is the desired right . 䉭 䉭ABC AB b Q F E C c C A B Q F E Geometry in the Real World In the manufacturing process, the parts of many machines must be congruent. The two sides of the hinge shown are congruent. A B E C D 1 2 Figure 3.21 EXAMPLE 4 Cite the reason why the right triangles and in Figure 3.21 are congruent if: a and b and C is the midpoint of c and d and bisects Solution a HL b AAS c ASA d SAS 쮿 BD EC AB ⬵ EC ∠1 ⬵ ∠2 BC ⬵ CD BD ∠A ⬵ ∠E AC ⬵ ED AB ⬵ EC 䉭ECD 䉭ABC Exs. 10–11 The square of the length c of the hypotenuse of a right triangle equals the sum of squares of the lengths a and b of the legs of the right triangle; that is, . c 2 = a 2 + b 2 PYTHAGOREAN THEOREM a A B C A 쮿 In applications of the Pythagorean Theorem, we often arrive at statements such as . Using the following property, we see that or . c = 5 c = 1 25 c 2 = 25 Technology Exploration Computer software and a calculator are needed. 1. Form a right with m C ⫽ 90⬚. 2. Measure AB, AC, and BC. 3. Show that AC 2 ⫹ BC 2 ⫽ AB 2 . Answer will probably not be “perfect.” ∠ 䉭 ABC Let x represent the length of a line segment, and let p represent a positive number. If , then x = 1p. x 2 = p SQUARE ROOTS PROPERTY The square root of p, symbolized represents the number that when multiplied times itself equals p. As we indicated earlier, because . When a square root is not exact, a calculator can be used to find its approximate value; where the symbol ⬇ means “is equal to approximately,” because . 21.9961 L 22 4.69 4.69 = 122 L 4.69 5 5 = 25 125 = 5 1p, EXAMPLE 5 Find the length of the third side of the right triangle. See the figure below. a Find c if and . b Find b if and . Solution a , so or . Then b , so or . Subtracting yields , so b = 151 L 7.14. b 2 = 51 100 = 49 + b 2 10 2 = 7 2 + b 2 c 2 = a 2 + b 2 c = 1 100 = 10. c 2 = 36 + 64 = 100 c 2 = 6 2 + 8 2 c 2 = a 2 + b 2 c = 10 a = 7 b = 8 a = 6 Exs. 12–14 C A a c b B Exercises 3.2 In Exercises 1 to 8, plan and write the two-column proof for each problem.

1. Given: and are

right Prove: 䉭ABC ⬵ 䉭ABD CA ⬵ DA ∠s ∠2 ∠1

3. Given: P

is the midpoint of both and Prove:

4. Given:

and Prove:

5. Given: and are

right Prove:

6. Given: and

Prove: 䉭RST ⬵ 䉭VST ∠3 ⬵ ∠4 ∠1 ⬵ ∠2 䉭RST ⬵ 䉭VST ∠1 ⬵ ∠2 ∠s ∠V ∠R 䉭MNP ⬵ 䉭RQP MN ⬵ QR MN 7 QR 䉭MNP ⬵ 䉭RQP NQ MR

2. Given: and are

right bisects Prove: 䉭ABC ⬵ 䉭ABD ∠CAD AB ∠s ∠2 ∠1 C B D A 1 2 Exercises 1, 2 M N P R Q Exercises 3, 4 S T R V 1 2

3 4

Exercises 5–8 쮿 Video exercises are available on DVD.