393 The line of centers of two externally tangent circles is tangent to the circle described upon their common external tangent as a diameter. [The tan gent at P passes through the center of O C'' (§ 254).] 394 If two circles are tangent externally, the cir cles described upon their common external tangents as diameters are tangent. [Ex. 393.] 395 H is the mid-point of the common external tangent of two tangent circles whose centers are C and C'. Prove that the angle CHC' is a right angle. 396 If two circles intersect, their common tangent subtends supplementary angles at the points of intersection. [To prove that SAT + SBT 2 rt. 4. Prove that SBT = 2 AST + 2 ATS.] அ. 397 Circles described on two sides of a triangle as diameters intersect on the third side or the third side produced. 398 If circles are described on the sides of a quadrilateral as diameters, the common chord of two consecutive circles is parallel to the common chord of the other two circles. [Draw the diagonals of the quad'l and apply Ex. 397.] C E D 399 If two circles intersect, any two parallel straight lines drawn through the points of intersection and terminated by the circumferences are equal. [ E and F are supplementary. .. CEFD is a .] 400 If two circles are internally tangent, the longest chord of the one tangent to the other is perpendicular to the line of centers. 401 If the diagonals of an inscribed quadrilateral intersect at the center of the circle, the figure is a right parallelogram. [ A=ZB=2C = Z D = a rt. ≤ (§ 277).] B 402 The chords of a circle bisecting the arcs subtended by the sides of an inscribed equilateral triangle meet those sides in points of trisection. 403 If three circles are externally tangent, two by two, their common internal tangents meet in a point. 404 If three circles are externally tangent, two by two, the inscribed circle of the triangle formed by joining their centers is the circumscribed circle of the triangle formed by joining their points of contact. 405 Radii which trisect a chord do not trisect the subtended arc. PROBLEMS OF CONSTRUCTION Thus far in the Geometry we have supposed the necessary figures constructed. We shall now show how to construct geometric figures with the straight edge and compasses, the only instruments allowed in Euclidean Geometry. PROPOSITION XIV. PROBLEM 283 To draw a perpendicular to a given line from a given point without the line. DATA. P is a given point without the line AB. REQUIRED. To draw a perpendicular to AB from P. SOLUTION From P as a center, and with a radius sufficiently great, describe an arc cutting AB at H and K. From H and K as centers, and with a radius greater than half of HK, describe arcs intersecting at C. Draw PC and produce it to meet AB at D. CONCLUSION. PD is the required. Q. E. F. PROOF. P and C are by construction each equidistant from H and K. .. PD is L to AB. § 188 Q. E. D. 284 To draw a perpendicular to a given line from a given point in the line. FIRST METHOD. DATA. P is a point in the line AB (Fig. 1). CONCLUSION. SOLUTION Take PH =PK. From H and K as centers, and with a radius greater than PH, describe arcs intersecting at C. Draw CP. CP is the perpendicular required. Q. E. F. PROOF. P and C are by construction each equidistant from H and K. SECOND METHOD. DATA. P is an end-point of the line SP (Fig. 2). SOLUTION From C, any point without SP, as a center, and with CP as a radius, describe an arc intersecting SP at D. produce it to meet the arc at E. Join EP. CONCLUSION. EP is the required. Draw DC and PROOF. Q. E. F. The angle P is a right angle. § 277 Q. E. D. .. EP is to SP. DISCUSSION. In the second method, the point C must be without the required perpendicular. PROPOSITION XVI. PROBLEM 285 To bisect a given straight line. X DATA. AB is a straight line. REQUIRED. To bisect AB. SOLUTION From A and B as centers, and with a radius greater than the half of AB, describe arcs intersecting at C and D. Join CD. Q. E. F. C and D are by construction each equidistant from PROPOSITION XVIII. PROBLEM 287 To bisect a given angle. B C DATA. ABC is a given angle. REQUIRED. To bisect the angle ABC. SOLUTION From B as a center, and with any convenient radius, describe an arc cutting the sides of the angle at D and E. From D and E as centers, and with a radius sufficiently great, describe arcs intersecting at O. Join OB. 288 SCHOLIUM. By the same construction, each half of the angle may be bisected. Hence an angle may be divided into 2, 4, 8, 16, 2", equal angles. EXERCISE 406. To trisect a right angle. [Let ABC be a rt. Z. From B as a center describe an A arc cutting the sides of the angle at A and C. From A and C as centers, and with AB as a radius, describe arcs cutting the arc AC at D and E. Then EB and DB trisect the For the ADB and CEB are equilateral by ABD = ≤ CBE = 60°. .. ≤DBC = < ABE = 30°.] ZABC. const... B NOTE. There is no method known in Geometry for trisecting any plane angle, and mathematicians believe that the problem is impossible. |