CASE II. When both of the parallels are secants. Conclusion. AC = BD. § 164 § 34 § 156 Suggestion. Draw EF parallel to AB and tangent at a point M. Apply Case I. CASE III. When both of the parallels are tangents. Hypothesis. AB is tangent at M and CD tangent at N; AB CD. Conclusion. Case I. Arc MEN = arc MFN. Draw secant EF parallel to AB. Apply Write the proof of the theorem in full. EXERCISES 1. Prove that the two non-intersecting chords which join the extremities of two parallel chords are equal. 2. Prove that the two intersecting chords which join the extremities of two parallel chords are equal. 3. Prove that an inscribed trapezoid is isosceles. 4. Prove that the diagonals of an inscribed trapezoid are equal. 5. Prove, by using Case III of the theorem in § 172, that a diameter of a circle bisects the circle. 6. Prove that if a tangent and a secant intercept equal arcs of a circle, they are parallel. 7. Prove that if two secants which do not intersect within the circle intercept equal arcs, they are parallel. 8. Prove that the opposite sides of an inscribed equilateral hexagon are parallel. B 173. The relations of angles and arcs. If a central angle AOB is divided into unit angles, such as degrees, by drawing radii, then AB is divided into an equal number of equal arcs. Why? If one of these equal arcs is taken as the unit of measure of arcs, then the numerical measure of AB is the same as that of ▲ AOB. This relation may be proved to be equally true when the units of measure are not con A tained exactly in ≤ AOB and AB, that is, when the numerical measures of ZAOB and AB can be expressed only approximately. The proof is omitted from this course. In general, if the unit of measure of arcs is chosen as the arc intercepted by a unit central angle, A central angle has the same numerical measure as its intercepted arc. 174. Corollary. In the same circle or equal circles, any two central angles have the same ratio as their intercepted arcs. This follows from the principle in § 173 by definition of ratio. See § 116. 175. Degrees, minutes, and seconds of arc. An arc which a central angle of one degree intercepts is called a degree of arc. Similarly, the arcs which central angles of one minute and of one second intercept are called a minute of arc and a second of arc, respectively. Hence, the number of degrees, minutes, and seconds of arc in an arc equals the number of degrees, minutes, and seconds in the central angle which intercepts it. Thus, a central angle of 62° 15' 30" intercepts an arc of 62° 15' 30". It follows that an entire circle equals 360°, a semicircle equals 180°, and a quadrant equals 90° of arc. In all work that follows, arcs will be measured in degrees, minutes, and seconds. Two arcs whose sum is 90° are called complementary arcs, and two arcs whose sum is 180° are called supplementary arcs. EXERCISES 1. Find the complement of an arc of 47° 24′ 40′′. 2. Find the supplement of an arc of 125° 52′ 15′′. 3. If a circle is divided into three arcs, two of which are 120° 30′ 15′′ and 110° 40' 30", respectively, find the number of degrees, minutes, and seconds in the third arc. 4. Prove that in the same circle or equal circles the complements of equal arcs are equal. 5. Prove that in the same circle or equal circles the supplements of equal arcs are equal. 6. How many degrees are there in the arc subtended by each side of an inscribed equilateral hexagon? Octagon? 7. If equilateral ▲ ABC is inscribed in a circle with center O, how many degrees are there in AB? In ▲AOB? 8. If AB is a side of an equilateral pentagon inscribed in a circle with center O, how many degrees are there in AB? In ▲ AOB? 9. Has a degree of arc the same length in two unequal circles? 176. Inscribed angles. An angle is said to be inscribed in a circle when its vertex is on the circle and its sides are chords, as ABC. Angle ABC intercepts AC. An angle is said to be inscribed in an arc when its vertex is on the arc and its sides pass through the extremities of the arc. Thus, ABC is inscribed in arc ABC. A 177. Theorem. An inscribed angle of a circle has the same measure as one half of its intercepted arc. CASE I. When the center is on one side of the angle. Hypothesis. BAC is inscribed in circle with center 0, and O is on AC. Conclusion. BAC has the same measure as BC. Proof. 1. BAC is inscribed and center O is on AC. Ax. V 7. .. ≤ BAC = { ≤ BOC. 8. But / BOC has the same measure as BC. 9. .. ≤ BAC has the same measure as § 173 BC. Ax. V CASE II. When the center is within the angle. Hypothesis. BAC is inscribed and center O is within ZBAC Conclusion. Suggestions. ▲ BAD and CASE III. Hypothesis. of Z BAC. <BAC has the same measure as 1⁄2 BC. Draw diameter AD. Apply Case I to Conclusion. BAC has the same measure as BC. Suggestions. Draw diameter AD. Apply Case I to Z DAB and DAC, and subtract. Write the proof of the theorem in full. 178. Corollary 1. An angle inscribed in a semicircle is a right angle. The proof is left to the student. 179. Corollary 2.- All angles inscribed in the same arc are equal. The proof is left to the student. B C E B EXERCISES 1. In the figure of Case I, § 177, if BC is one fourth of the entire circle, how many degrees in BAC? 2. If an angle inscribed in an arc of a circle is a right angle, the arc is a semicircle. 3. An angle inscribed in an arc less than a semicircle is obtuse. 4. An angle inscribed in an arc greater than a semicircle is acute. 5. If an angle inscribed in an arc is obtuse, the arc is less than a semicircle. 6. If an angle inscribed in an arc is acute, the arc is greater than a semicircle. 7. Prove that the center of any circular object may be located by means of a carpenter's square as follows: Lay the square on the object, with the heel at the rim, and mark the points A and B where the blades cross the rim. Now, by placing the blade of the square on A and B, find the middle point of AB. That is the center. B |