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Let the diverging lines EA, EC, EB, and ED terminate in the harmonic section of the straight line AD; any intercepted straight line FG will be likewise cut by them harmonically, or FG : GI :: FH: HI. For, through the points B and I, draw (I. 23.) KL and MN parallel to AE.

Because the parallels AE and BL are intercepted by the diverging lines DA and DE, AD : DB :: AE : BL (VI. 2.); and for the same reason, the parallels AE and BK being intercepted by the diverging lines AB and EK, AC : CB : : AE : BK. And since AD is divided harmonically, AD : DB:: AC : CB; wherefore AE: BL: : AE : BK, and consequently(V. 8. and 4.) BL = BK. But, KL being parallel to MN, BL: BK :: IN: IM (VI. 2. cor. 2.); consequently, BL being equal to BK, IN must also be equal to IM; whence FE : IN : : FE: IM. Again, FE: IN :: FG : GI, for the parallels FE and IN are cut by the diverging lines GF and GE ; and FE: IM : : FH: HI, since the parallels FE and IM are cut by the diverging lines FI and EM. Wherefore, by identity of ratios, FG : GI: ; FH: HI; or the intercepted straight line FG is cut harmonically in the points H and I. -

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A straight line drawn from the concourse of two tangents to the concave circumference of a circle, is divided harmonically, by the convex circumference and the chord which joins the points of contact. - -

Let ED and FD be two tangents applied to the circle AEBF; the secant DA, drawn from their point of congourse, will be cut in harmonic proportion, by the convex circumference EBF and the chord EF which joins the points of contact,-or AD : DB :: AC : CB. For the tangents ED and FD are equal (III. 22, cor), and EDF being thus an isosceles triangle, DE” = DC + EC.CF (II. 20.); (but III. 26. cor. 2.) DE” is also equal to AD.DB, and the chords AB and EF, by their mutual intersection, make the rectangle EC, CF equal to AC, CB. Whence DC*- AD.DB—AC.CB, and therefore (VI. 7. cor.) AC : CB: ; AD : DB.

Cor. Hence by applying Prop. 7, it follows, that half the chord AB is a mean proportional between the distances of its middle point from C and D; and that, when AD passes through the centre of the circle, the square of the radius is equivalent to the rectangle under the distances of the chord and of the intersection of the tangents from the Centre.



A straight line which bisects, either internally or externally, the vertical angle of a triangle, will divide its base into segments, internal or external, that are proportional to the adjacent sides of the triangle. * -

Let the straight line BD bisect the vertical angle of the triangle ABC ; it will cut the base AC into segments which have the same ratio as the adjacent sides, or AD : DC : : AB : BC.

For through C draw CE parallel to DB (I. 23.), and meeting the production of AB in E.

Because DB and CE are parallel, the exterior angle ABD is equal to BEC, and the alternate angle DBC equal to IC BCE (I. 22.); wherefore the angle ABD being equal by hy- . pothesis to DBC, the angle IB BEC is equal to BCE, and con- sequently (I, 11.) the triangle CBE is isosceles, or BE is equal to BC. But the parallels DB and CE cut the diverging lines AC and AE proportionally (VI. 1.), or AD: DC : : AB : BE; that is, since BE= BC, AD : DC: ; AB: BC.

Again, let the vertical line BD bisect the exterior angle CBG of the triangle ; it will divide the base into external segments AD and DC, which are also proportional to the adjacent sides AB and BC.


For through C draw CE parallel to DB, and meeting AB in E.

The equal angles GBD and DBC are, from the properties of parallel straight lines, respectively equal to BEC. and BCE, and consequently the triangle CBE is isosceles, or the side BC is equal to BE. And since the diverging lines AD and AB are cut by the parallels DB and CE proportionally, AD : DC : : AB : BE or BC.

Cor. Hence the converse of the Proposition is likewise true, or if a straight line be drawn from the vertex of a triangle to cut the base in the ratio of the adjacent sides, it will bisect the vertical angle; for it is evident, from

VI. 6. cor., that a straight line is only capable of a single section, whether internal or external, in a given proportion. Scholium. The vertical line BD must bisect the base AC of the triangle, when the sides AB and BC are equal. In the case where BD bisects the exterior angle CBG, if AB be supposed to approach to an equality with BC, the straight line EC will come nearer to AC, and consequently the incidence D of the parallel BD with AC will be thrown continually more remote. But when the side AB is equal to BC, the straight line BD, being now parallel to AC, will never meet it, or there can be no equality of external section; for though the ratio of AD to CD tends towards the ratio of equality as the point D retires, yet the


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constant difference AC between those distances must always bear a sensible relation to them. After BD, in turning about the point B, has passed the limits of distance beyond C, it re-appears in an opposite direction beyond A, when AB, receding from equality, has become less than BC.

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Triangles are similar, which have their corresponding angles equal.

Let the triangles ABC and DEF have the angle CAB equal to FDE, CBA to FED, and consequently (I. 30.) the remaining angle BCA equal to EFD ; these triangles are similar, or the sides in both which contain equal angles are proportional.

For make BG equal to ED, and draw GH parallel to AC. -

Because GH is parallel to AC, the exterior angle BGH is equal (I. 22.) to BAC, that is to EDF ; and the angle at B is, by hypothesis, equal to that at E, and the interjacent AIT C ID F side BG was made equal to ED; wherefore (I.20.) the triangle GBH is equal to DEF. But, the diverging lines BA and BC being cut proportionally by the parallels AC and GH (VI. I.), AB

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