368 If two equal chords of a circle intersect, the segments of one are equal respectively to the segments of the other. 369 AS and AT are fixed tangents to a circle, and BC is a third tangent meeting AS at B and AT at C. Prove that the perimeter of the triangle ABC is constant for all positions of BC. the perimeter = AS + AT.] [Show that B P T 370 The shortest and the longest line that can be drawn from a given point to a given circumference lie in the diameter that passes through the point. 371 Two circles intersect at A and B. X and Y are any two points on one circumference. XA and XB produced meet the other circumference at P and T; YA and YB produced meet it at S and Q. Prove ST and PQ parallel. [<XAY = ≤ XBY. Z SAP = < TBQ. ... arc SP =arc TQ. ... ST is to PQ.] B T P 372 If two circles are tangent internally, and the radius of the first equals the diameter of the second, any chord of the first drawn from the point of contact is bisected by the circumference of the second. 373 If two secants are drawn through the point of contact of two tangent circles, the chords of the intercepted arcs are parallel. Two cases. [<C = ▲ APS = Z BPT < BPT = 2 D.] IS A P 374 If a side of an inscribed equilateral triangle is perpendicular to a radius it bisects the radius. 375 The altitude of an inscribed equilateral triangle is three fourths of the diameter. [Ex. 151.] X A Y 376 Two circles intersect at A and B; XAY and PBQ are any two straight lines terminated by the circumferences. Prove PX and QY parallel. [<X sup. of ≤ ABP, and ≤Y = sup. of ≤ ABQ (§ 279).] 377 Circles inscribed in and circumscribed about a square are concentric. 378 If two circles intersect, the straight line drawn through a point of intersection parallel to the line of centers and terminated by the circumferences is twice the line of centers. 379 If two circles intersect, the longest line which can be drawn through a point of intersection and terminated by the circumferences is the one parallel to the line of centers. 380 Two circumferences cannot bisect each other. 381 If the vertical angle of an isosceles triangle inscribed in a circle is twice an angle at the base, the base is a diameter of the circle. 382 If two triangles inscribed in a circle are mutually equiangular, they are also mutually equilateral. A H 383 Is Ex. 382 true if both triangles are circumscribed about the circle ? 384 Two circles intersect at A and B. PAH and PBK are straight lines terminated by the circumfer- P Prove that the tangent at P is parallel to HK. [< H = 2 ABP (Ex. 289) = ≤ APT (§§ 275, 281).] ences. 385 The altitude of an equilateral triangle is three times the radius of the inscribed circle. 386 The radius of the circle circumscribed about an equilateral triangle is twice the radius of the inscribed circle. 387 The diameter of a circle inscribed in a right triangle is equal to the difference between the sum of the legs and the hypotenuse. [§ 254.] 388 If a circle is inscribed in a right triangle, and another is circumscribed about it, the sum of the legs of the triangle is equal to the sum of the diameters of the two circles. [§ 277, Ex. 387.] 389 A circle inscribed in the triangle ABC touches AB at E and AC at D. Prove that the circle inscribed in the triangle AED has the midpoint of the arc ED as its center. 390 If a triangle is circumscribed about a circle, the triangle formed by joining the points of contact is acute. [Z ADF Z AFD. .. Z ADF is acute (§ 154). .. ▲ DEF is acute.] www. A D 391 Two circles are tangent externally at P; S and T are the points of contact of their common external tangent. Prove that the angle SPT is a right angle. [Draw the common tangent HP. The ▲ HSP and HTP are isosceles (§ 254).] 392 If two circles are tangent externally, a circumference described upon their common external tangent B A D E F as a diameter passes through the point of contact. [Ex. 391.] с с 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 circles 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 inter[To prove that ≤ SAT + ≤ SBT = 2 rt. . section. Prove that SBT = < AST + 2 ATS.] H Τ A B D 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.] 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 □.] E B D 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 = 2 C = 2D = ▲ 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. 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. PROPOSITION XV. PROBLEM 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). SOLUTION Take PH = PK. From H and K as centers, and with a radius greater than PH, describe arcs intersecting at C. CONCLUSION. 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 PROOF. Draw DC and The angle P is a right angle. § 277 Q. E. D. .. EP is 1 to SP. DISCUSSION. In the second method, the point C must be without the required perpendicular. |