BOOK IV. PROBLEMS. IN this section, we have, in most instances, merely shown the construction of the problem, and referred to the theorem or theorems that the student may use, to prove that the object is attained by the construction. In obscure and difficult problems, however, we have gone through the demonstration as though it were a theorem. PROBLEM I. To bisect a given finite straight line. Let AB be the given line, and from its extremities, A and B, with any radius greater than one half of AB, (Postulate 3), describe arcs, cutting A each other in n and m. Draw the line nm; and C, where it cuts AB, will be the middle of the given line. Proof, (B. I, Th. 18, Sch. 2). PROBLEM II. To bisect a given angle. Let ABC be the given angle. With any radius, and B as a center, describe the arc AC. From A and C, as centers, with a radius greater than one half of AC, describe arcs, intersecting in n; join B and n; the joining line will bisect the given angle. Proof, (Th. 21, B. I). PROBLEM III. From a given point in a given line, to draw a perpendicular to that line. Let AB be the given line, and C the given point. Take n and m, at equal distances on opposite sides of C; and with the points m and n, as centers, and any radius greater than no or mC, describe An arcs cutting each other in S. Draw m B SC, and it will be the perpendicular required. Proof, (B. I, Th. 18, Sch. 2). The following is another method, which is preferable, when the given point, C, is at or near the end of the line. Take any point, O, which is manifestly one side of the perpendicular, n as a center, and with OC as a radius, describe a circumference, cutting AB in m and C. Draw mn through the points m and 0, and meeting the arc again in n; mn is then a diameter to the circle. Draw Cn, and it will be the perpendicular required. Proof, (Th. 9, B. III). PROBLEM IV. From a given point without a line, to draw a perpendicular to that line. Let AB be the the given point. oblique line, as Cn. given line, and C From C draw any Find the mid dle point of Cn by Problem 1, and with that point, as a center, describe a semicircle, having Cn as a diameter. From m, where this semi-cir n m cumference cuts AB, draw Cm, and it will be the perpen dicular required. Proof, (Th. 9, B. III). PROBLEM V. At a given point in a line, to construct an angle equal to a given angle. Let A be the point given in the line AB, and DCE the given angle. With Cas a center, and any radius, C CE, draw the arc ED. With A as a center, and the radius AF-CE, describe an indefinite arc; and with F as a center, and FG as a radius, equal to ED, describe an arc, cutting the \G A B other arc in G, and draw AG; GAF will be the angle required. Proof, (Th. 2, B. III). PROBLEM VI. From a given point, to draw a line parallel to a given line. Let A be the given point, and BC the given line. Draw AC, making an angle, ACB; and from the given point, A, in the line AC, draw the angle CAD= ACB, by Problem 5. Since AD and BC make the same angle with AC, they are, therefore, parallel, (B. I, Th. 7, Cor. 1). PROBLEM VII. To divide a given line into any number of equal parts. Let AB represent the given line, and let it be required to di- B viders, as Aa, and set it off on the line AD, thus making the parts Aa, ab, bc, etc., equal. Through the last point, e, draw EB, and through the points a, b, c, and d, draw parallels to eB, by Problem 6; these parallels will divide the line as required. Proof, (Th. 17, Book II). PROBLEM VIII. To find a third proportional to two given lines. To find a fourth proportional to three given lines. Let AB, AC, AD, represent the hrce given lines. Place the first two at any angle, as BAC, and draw BC On AB place AD, and from the point D, draw DE parallel to BC, by Problem 6; AE will be the fourth proportional required. Proof, (Th. 17, B. II). A A A E PROBLEM X. -D To find the middle, or mean proportional, between two given liner B B Place AB and BC in one right ine, and on AC, as a diameter, de- A scribe a semicircle, (Postulate 3), B and from the point B, draw BD at right angles to AC, (Problem 3); BD is the mean proportional required. Proof, (B. III, Th. 17, Cor.). PROBLEM XI. A B To find the center of a given circle. Draw any two chords in the given circle, as AB and CD, and from the middle points, m and n, draw perpendiculars to AB and CD; the point at which these two perpendiculars intersect will be the center of the circle. Proof, (B. III, Th. 1, Cor.). PROBLEM XII. A To draw a tangent to a given circle, from a given point, either in or without the circumference of the circle. When the given point is in the circumference, as A, draw the radius AC, and from the point A, draw AB perpendicular to AC; AB is the tangent required. Proof, (Th. 4, B. III). Di When the given point is without the circle, as A, draw AC to the center of the circle; on AC, as a diameter, describe a semicircle; and from B, where the semi-circumference cuts the given circumference, draw AB, and it will be tangent to the circle. Proof, (Th. 9, B. III), and, (Th. 4, B. III). |