426 I'HILOSOPHICAL TRANSACTIONS. [aNNO I67O. 



As db : bg : ; radius : tang, bdg or cdli ; 

 and cb : bg : I radius : tang, beg or aci ; 

 therefore db X tang, cdh =z= bg X rad. = cb X tang, del, 

 and db : cb : : tang, dci : tang. cdh. 

 That is, db will be to cb, as the tangent of half the mean anomaly, to the 

 tang, of half the true anomaly^ consequently, by the preceding rule, as the 

 greatest distance of the planet is to its least distance. Therefore db will be 

 equal to the greatest distance, ch the least, and ah the eccentricity. 



And since the same demonstration holds of all the other points of inter- 

 section, viz. that the perpendiculars from them to the line cd fall upon the point 

 h\ the line joining these intersections must coincide with the perpendicular 



III. Having drawn the diameter hah, take the arch hi equal to the arch id, 

 and draw kc and hi intersecting each other in p; let fall the perpendicular hr 

 from h upon bgf, and parallel to the line of the apses cd\ then the angle rhs 

 will be equal to the half difference of ch the true anomaly and di the mean. 

 Then draw the right line A(3 from the same point h, making with ^A an angle 

 equal to rhs, and meeting the line of the apses in |3, so shall the angle ^ahhe. 

 the measure of the arch ch, or of the true anomaly, and (3Aa the half difference 

 of the true and mean anomalies, by the construction; also the external angle 

 c^h, which is equal to the two internal and opposite angles |3aA and ^ha, and 

 consequently compounded of the true anomaly, and half its difference from the 

 mean, will be half the sum of the true and mean anomalies. Consequently, by 

 the former analogy of the first corollary, as the sine of cj3A, is to the sine of 

 |3Aa, so is the radius ah, to the eccentrity a|3. But it was demonstrated above, 

 that ah IS also equal to the eccentricity. Therefore the point |3 coincides with 

 the point h. 



Then erecting ht perpendicular to hh, this, if produced, will fall on the point 

 of intersection p. For the triangles rhs and hht are similar by the construction ; 

 as also the triangle hpk is similar to hgi, for the angles j&j^ A and gih are equal, 

 insisting upon the same arch, ch, as also the angles j&A^ and ghi insisting on the 

 equal arcs hi and id, consequently the two other angles hpk and hgi are equal ; 

 and taking away the equal angles hht and rhs from the equals phk and ghi, 

 there will remain phh equal to ghr. Whence I argue thus, srli = thh, and 

 rhs =. hht, therefore hsr z= hth, consequently their complements to a semi- 

 circle are equal, viz. rsi = htk, and sig = tkp, therefore also igs = kpt, which 

 subtracted from the equal angles igh and kph, there remains hgs = hpt, and 

 ghr = phh; therefore also hrg = hbp; but hrg is a right angle, consequently 

 hhp is also a right angle; and since hht is a right angle by construction, th will 



