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(a) By Principle 1, mj P 90 . Then mj A x 18. Hence, m j A 36 . mj O 90 . If mj A 2x and mj O 3x, then 5x 90 and

Fig. 6-26



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(b) By Principle 1, mj P mj Q 90 . Since mj P mj Q mj O mj A Since mj O 140 , mj A 40 . By Principle 5, mj 1 1 mj A 20 . 2 (c) By Principle 1, m j DPQ mj PQC of ^ PBQ, mj 3 90 60 150 . 90 . Since mj 1 30 , mj 2

6.7 Applying principle 4 (a) AP, BQ, and AB in Fig. 6-27(a) are tangents. Find y. (b) ^ ABC in Fig. 6-27(b) is circumscribed. Find x. (c) Quadrilateral ABCD in Fig. 6-27(c) is circumscribed. Find x.

Fig. 6-27

Note!





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(a) By Principle 4, AR (b) By Principle 4, PC 11 8 19. (c) By Principle 4, AS 10 8 18. 6, and RB 8, QB 10, CR y. Then RB 4, and AP 5, and RD AB AR AQ. Then AQ SD. Then RD 14 6 8. Hence, y RB 8. AP AS PC SD

6.8 Finding the line of centers Two circles have radii of 9 and 4, respectively. Find the length of their line of centers (a) if the circles are tangent externally, (b) if the circles are tangent internally, (c) if the circles are concentric, (d) if the circles are 5 units apart. (See Fig. 6-28.)

Fig. 6-28

Let R radius of larger circle, r 9 and r 9 and r 4, OO9 4, OO9 radius of smaller circle. R R r r 9 9 4 4 13. 5.

(c) Since the circles have the same center, their line of centers has zero length. (d) Since R 9, r 4, and d 5, OO9 R d r 9 5 4 18.

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In the preceding example you de ned three child elements that were required to occur in a speci c sequence. Now you want to allow the author a choice between three elements. The DTD equivalent is

Prove: Tangents to a circle from an outside point are congruent (Principle 4). Given: Circle O AP is tangent at P. AQ is tangent at Q. To Prove: AP > AQ Plan: Draw OP, OQ, and OA and prove ^ AOP > ^ AOQ.

PROOF:

Three pieces of information are needed to calcu late depreciation on any asset: total cost, estimat ed useful life, and residual (or scrap) value.

Statements 1. Draw OP, OQ, and OA. 2. 3. 4. 5. 6. OP > OQ j P and j Q are right angles. OA > OA ^ AOP > ^AOQ AP > AQ

Reasons 1. A straight line may be drawn between any two points. 2. Radii of a circle are congruent. 3. A tangent is to radius drawn to point of contact. 4. Reflexive property 5. Hy-leg s 6. Corresponding parts of congruent ^ are congruent.

A central angle has the same number of degrees as the arc it intercepts. Thus, as shown in Fig. 6-29, a central angle which is a right angle intercepts a 90 arc; a 40 central angle intercepts a 40 arc, and a central angle which is a straight angle intercepts a semicircle of 180 . Since the numerical measures in degrees of both the central angle and its intercepted arc are the same, we may restate the above principle as follows: A central angle is measured by its intercepted arc. The symbol may be used to mean is measured by. (Do not say that the central angle equals its intercepted arc. An angle cannot equal an arc.)

Fig. 6-29

An inscribed angle is an angle whose vertex is on the circle and whose sides are chords. An angle inscribed in an arc has its vertex on the arc and its sides passing through the ends of the arc. Thus, j A in Fig. 6-30 is an inscribed angle whose sides are the chords AB and AC. Note that j A intercepts BC and is inscribed in BAC.

Fig. 6-30

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