Conversely :-If through an end of a chord of a circle a straight line be drawn making angles with the chord equal to the angles in the alternate segments, this straight line shall be a tangent to the circle. Let BD be a chord of the circle ABCD, and let EBF be drawn through (and .. EBD : = in segment BCD), Draw the diamr. BA through B. Join AD and produce it to F. Then BAD = 4 FBD. But BFD is common to as ABF, BDF, 3rd ABF=3rd 4 BDF, =a rt. 4; ... EBF is the tangent at B. This converse can also be demonstrated indirectly. It is important for subsequent work. Ex. 362.—The tangent at A to the circum-circle of the triangle ABC is parallel to any anti-parallel to BC with respect to A. (See Ex. 105 and 257.) Ex. 363.-ABC is a triangle; D and E are taken on AB, AC such that DE is parallel to BC: show that the circum-circles of triangles ABC, ADE have a common tangent at A. Ex. 364.-ABCD is a cyclic quadrilateral; AD, BC produced meet in E prove that the tangent at E to the circum-circle of CDE is parallel to AB. Ex. 365.-Use III. 32 to show that the tangents to a circle from an external point are equal. Ex. 366.-Two circles touch each other (externally or internally); show that : (i.) If through the point of contact any straight line be drawn it will cut off similar segments. (Draw the common tangent at the point of contact.) (ii.) If through the point of contact any straight line be drawn to cut the circles again, the tangents at the other points of section will be parallel. (iii.) If through the point of contact any two straight lines be drawn to cut the circles again, the chords joining the other points of section will be parallel. PROPOSITION 33. PROBLEM. Upon a given straight line to describe a segment of a circle containing an angle equal to a given rectilineal angle. Let AB be the given st. line, and C the given rect. : it is reqd. to describe on AB a segt. of a containing an equal to C. Bisect AB at F. (1) If C is a rt. 4, with centre F and rad. FA or FB, describe the semi-O АНВ. Then AHB is a rt. 4, (2) If .. AHB=LC. C be not a rt. —, at A make BAD equal to C, and draw AG ir to AD; through F draw FG Lr to AB, and join GB. and rt. AFG=rt. BFG, ... AG=BG, .. AHB described with centre G and rad. GA shall pass through B. Note that we are shown how to describe a circle which shall touch a given straight line, at a given point, and pass through another given point. Ex. 367. To find a point O within a triangle ABC such that angle OAB = angle OBC = angle OCA. B Describe a passing through C, and touching AB at A. Draw the chord AP parallel to BC. Join BP, cutting the Oce in O. Ex. 368.—Find a point O' within a triangle ABC, such that angle O'BA= angle O'’CB = angle O’AC. These two points O and O' are called the Brocard points of the triangle ABC. The construction given above we owe to Mr. R. F. Davis, M.A. PROPOSITION 35. THEorem. If two straight lines within a circle cut one another, the rectangle contained by the segments of one of them is equal to the rectangle contained by the segments of the other. Let the two chds. AC, BD of the ABCD intersect at E, then rect. AE, EC=rect. BE, ED. B (1) If each passes through the centre, E is the centre, rect. AE, EC = rect. BE, ED (·.· AE = EC = ED = EB). (2) If one of them, BD, pass through the centre F, and cut the other, AC, which does not pass through the centre, (3) If BD pass through the centre F, and cut AC, which does not pass through the centre, but not at rt. s, draw FG r to AC and join FC. Then AG=GC, and rect. BE, ED with sq. on EF = sq. on FB, [II. 5. = sq. on FC (·.· FB=FC). =sqs. on FG, GC, [I. 47. =sqs. on FG, GE with rect. AE, EC, [II. 5. But sq. on EF = sqs. on FG, GE, [I. 47. .. rect. BE, ED = rect. AE, EC. (4) If neither pass through the centre F, join EF and produce it to cut the Oce in H and G. H Then rect. BE, ED = rect. GE, EH, =rect. AE, EC, } by cases (2), (3). Ex. 372.-Prove the converse of III. 35. (Use the indirect method.) Ex. 373.—ABC is a triangle; AD, BE ir to BC, AC respectively. Cut in O. Show that rectangle AO, OD=rectangle BO, OE. |