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Join EB, and draw EO and BO to the centre O. The triangles EOD and BOD, having the side EO equal to

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are (I. 3. El.) also equal, and therefore EB is perpendicular to the diameter FH. Wherefore (VI. 9. El.) FA: AH :: FD: DH; but the ratio of FA to AH being given, and consequently that of FD to DH, the point D (VI. 6. El.) is given.


Make (VI. 3. El.) OA : OH :: OH: OD, and then D is the point required. For join OC and OB. Because OH = OC, OA: OC :: OC: OD; wherefore the triangles AOC and COD, having thus the sides about their common angle DOC proportional, are similar; and hence the angle OCA is equal to ODC. In the same manner, it is proved that the angle OBA is equal to ODB. But BOC being an isosceles triangle, the angle OCA is equal to OBA; whence the angle ODC is equal to ODB.

This porism is likewise derived from the local theorem given in Prop. 14. For AC, DC, and AB, DB being inflected in the same ratio, AC: AB :: DC: DB; and consequently (VI. 11. cor. El.) the angle BDC is bisected by DA.


A point being given in the circumference of a circle, another point may be found, so that two straight lines inflected from them to the opposite circumference, shall cut off, on a given chord, extreme segments, whose alternate rectangles shall have a given ratio.

Let the circle ADBE, the point A, and the chord DE, be given in position,-another point C may be found, such that straight lines AB and CB inflected to the opposite circumference, shall form segments containing rectangles DG, FE, and DF, GE, in the ratio of KM to LM.


Join CA, and produce it to meet the extension of the chord ED in H.

Because KM: LM:: DG. FE: DF. GE, by division KL: LM:: DG. FE-DF.GE: DF.GE; but DG.FEDF.GE=(DF+FG) (GE+FG)

DF.GE FG.DE, and consequently KL : LM ::

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FG. DE DF. GE. Make

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El.) AF.FB. Wherefore FH: FB:: AF: FG, and (VI.

15. El.) the triangles AFH and GFB are similar, and consequently the angle AHF is equal to FBG; but the angle AHF is given, since the points A, H, and D are given, and, therefore, the chord AC, cutting off from the given circumference, a segment that contains a given angle ABC or FBG is given, and thence the point C.


Produce the chord ED to H in the ratio of KM to LM, join HA, and, at any point B in the circumference, make the angle ABC equal to AHF; C is the point required.

For, the triangles AFH and GFB being evidently similar, FH: FB :: AF: FG, and FH.FG=FB.AF=DF.FE; whence FH.FG-DF.FG-DF.FE-DF.FG, or FGXDH =DF × GE. But KL: LM :: DE : DH :: FG × DE : FGX DH, and therefore KL: LM:: FGX DE: DFXGE; consequently (V. 9. El.) KM : LM :: FG × DE+ DF×GE, or DGX FE: DFXGE.

The porism now investigated arises naturally out of this problem :-" From two given points A and C, one of which lies in a given circumference, to inflect straight lines AB and CB, so as to intercept on the chord DE segments that contain rectangles DG, FE and DF, GE, which are in a given ratio." For, the point H being assumed as before, the analysis requires that the angle ABC should be made equal to AHF. Whence, if on AC, a segment of a circle were described containing that angle, its contact or intersection with the given circumference, would determine the point of inflection. Supposing, therefore, the two circles entirely to coincide, the problem will in that case become indeterminate, or admit of innumerable answers.


Two points and two diverging lines being given in position, straight lines, inflected from those points to one of the diverging lines, intercept segments, on the other, from points that may be found, and containing a rectangle which will be likewise assign


Let DF and EF be inflected, from the points D and E, to the diverging line AC; they will cut off segments, on AB, from points I and K which may be found, so that the rectangle IH, GK shall be given,


Join El and EA, DA and DK, and produce ED to meet AC in P. Since A, F, and P are so many points of inflection, it is evident, from the hypothesis, that IA.AK=IH.GK IN.NK; whence IH: IA

:: AK: GK, and, by division, AH: IA :: AG : GK, and alternately AH: AG :: IA: GK. Through E, draw LEM parallel to AB and meeting AC and FD produced; then (VI. 2. El.) LE: LM :: AH: AG ::IA: GK. Again, because IA. AK = IN. NK,

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IN:IA::AK:NK, by division ANIA :: AN: NK and consequently IA NK Wherefore, by substitution, LE: LM:: NK: GK, and LE: EM:: NK: GN, or alternately

Join DO.

LE: NK EM: GN, that is, (VI. 2. El.) ED: DN; hence (VI. 16. El.) the triangles LDE and KDN are similar, and LDK forms one single straight line. Since IA NK, LE: IA :: LE: NK, that is (VI. 2. El.) EO: OL:: ED: DN, and therefore (VI. 1. cor. 1. El.) DO is parallel to AB. But the parallels OD and LM being given in position, the points O and L, and thence I and K, are given, and consequently the rectangle IA, AK is given.


Draw DO, EL parallel to AB and meeting the extension of AC, join EO, LD, and produce them to meet AB in I and K; these are the points required. For DF and EF being inflected, LE: IA :: OE: OI :: ED: DN :: DM : DG :: LM: GK, and alternately LE: LM:: IA: GK; but LE: LM :: AH : HG, and therefore IA: GK :: AH: AG; consequently (V. 8. and 11. El.) IA: IH :: GK : AK, and IA.AK IH.GK.

The porism thus investigated follows from this problem : "Two straight lines AB and AC being given in position, with the points I and K, E and D, to find a point F, such that the inflected lines EF and DF shall intercept segments IH and GK, containing a given space:" For, when the points I and K have the position before assigned, the construction becomes indeterminate.


Three diverging lines being given in position, a fourth may be found, such that straight lines can be drawn intersecting all these and divided by them into proportional segments.

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