B Ᏼ . C B C 366. FIG. 1. FIG. 2. FIG. 3. Cor. I. All angles inscribed in the same segment are equal. (See Fig. 1.) 367. Cor. II. Any angle inscribed in a semicircle is a right angle. (See Fig. 2.) 368. Cor. III. The locus of the vertex of a right triangle having a given hypotenuse as base is the circumference having the same hypotenuse as diameter. 369. Cor. IV. Any angle inscribed in a segment less than a semicircle is an obtuse angle. (See Fig. 3.) 370. Cor. V. Any angle inscribed in a segment greater than a semicircle is an acute angle. (See Fig. 3.) Ex. 480. If an inscribed angle contains 24 angle degrees, how many arc degrees are there in the intercepted arc? how many in the rest of the circumference? Ex. 481. If an inscribed angle intercepts an arc of 70°, how many degrees are there in the angle ? Ex. 482. How many degrees are there in an angle inscribed in a segment whose arc is 140° ? Ex. 483. Construct any segment of a circle so that an angle inscribed in it shall be an angle of: (a) 60°; (b) 45°; (c) 30°. Ex. 484. Repeat Ex. 483, using a given line as chord of the segment. How many solutions are there to each case of Ex. 483 ? how many to each case of Ex. 484 ? Ex. 485. The opposite angles of an inscribed quadrilateral are supplementary. Ex. 486. If the diameter of a circle is one of the equal sides of an isosceles triangle, the circumference will bisect the base of the triangle. Ex. 487. By means of a circle construct a right triangle, given the hypotenuse and an arm. Ex. 488. By means of a circle construct a right triangle, given the hypotenuse and an adjacent angle. Ex. 489. Construct a right triangle, having given the hypotenuse and the altitude upon the hypotenuse. 371. Historical Note. Thales (640-546 B.C.), the founder of the earliest Greek school of mathematics, is said to have discovered that all triangles having a diameter of a circle as base, with their vertices on the circumference, have their vertex angles right angles. Thales was one of the Seven Wise Men. He had much business shrewdness and sagacity, and was renowned for his practical and political ability. He went to Egypt in his youth, and while there studied geometry and astronomy. The story is told that one day while viewing the stars, he fell into a ditch; whereupon an old woman said, "How canst thou know what is doing in the heavens, when thou seest not what is at thy feet?" According to Plutarch, Thales computed the height of the Pyramids of Egypt from measurements of their shadows. Plutarch gives a dialogue in which Thales is addressed thus, "Placing your stick at the end of the shadow of the pyramid, you made by the same rays two triangles, and so proved that the height of the pyramid was to the length of the stick as the shadow of the pyramid to the shadow of the stick." This computation was regarded by the Egyptians as quite remarkable, since they were not familiar with applications of abstract science. The geometry of the Greeks was in general ideal and speculative, the Greek mind being more attracted by beauty and by abstract relations than by the practical affairs of everyday life. PROPOSITION XX. PROBLEM 372. To construct a perpendicular to a given straight line at a given point in the line. (Second method. For another method, see § 148.) A B Given line AB and P, a point in AB. To construct a L to AB at P. I. Construction 1. With o, any convenient point outside of AB, as center, and with OP as radius, construct a circumference cutting AB at P and Q. 2. Draw diameter QM. 3. Draw PM. 4. PM is AB at P. II. The proof and discussion are left as an exercise for the student. The method of § 372 is useful when the point P is at or near the end of the line AB. Ex. 490. Construct a perpendicular to line AB at its extremity B, without prolonging AB. Ex. 491. Through one of the points of intersection of two circumferences a diameter of each circle is drawn. Prove that the line joining the ends of the diameters passes through the other point of intersection. PROPOSITION XXI. PROBLEM 373. To construct a tangent to a circle from a point outside. P X Given circle O and point P outside the circle. 1. Draw PO. I. Construction 2. With Q, the mid-point of PO, as center, and with Q0 as radius, construct a circumference intersecting the circumference of circle o in points T and V. 3. Draw PT and PV. 4. PT and PV are tangents from P to circle 0. II. The proof and discussion are left as an exercise for the student. HINT. Draw OT and OV and apply § 367. Ex. 492. Circumscribe an isosceles triangle about a given circle, the base of the isosceles triangle being equal to a given line. What restriction is there on the length of the base? Ex. 493. Circumscribe a right triangle about a given circle, one arm of the triangle being equal to a given line. What is the least length possible for the given line, as compared with the diameter of the circle? Ex. 494. If a circumference M passes through the center of a circle B, the tangents to B at the points of intersection of the circles intersect on circumference M. Ex. 495. Circumscribe an isosceles triangle about a circle, the altitude upon the base of the triangle being given. Ex. 496. Construct a common external tangent to two given circles. To construct a common external tangent to circles MTR and NVS. I. Construction 1. Draw the line of centers OQ. 2. Suppose radius OM > radius QN. Then, with with OLOM - QN as radius, construct circle LPK. 3. Construct tangent QP from point Q to circle LPK. as center and § 373. 4. Draw OP and prolong it to meet circumference MTR at T. 5. Draw QVOT. § 188. 6. Draw TV. 7. TV is tangent to circles MTR and NVS. II. The proof and discussion are left as an exercise for the student. Q toward O in the preceding figure, show when there are four common tangents; when only three; when only two; when only one; when none. |