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PROP. XXVII. THEOR.
If on the hypotenuse of a right-angled triangle, a rectangle of any altitude be described, and straight lines be drawn from the right angle to the remote angles of the rectangle, dividing the hypotenuse into three parts ; the rectangle of the extreme parts is to the square of the middle one, as the square of the altitude of the rectangle, is to the square of the hypotenuse.
On BC, the hypotenuse of the right-angled triangle ABC, let the rectangle BDEC be described, and let AD, AE be joined, cutting BC in F and G: BF.GC: FG2 : : BD2 : BC%.
Produce AB, AC, and let them meet DE produced in H, K. Then (VI. 21.) the triangles HDB, CEK are similar to one another, each of them being evidently similar to the triangle HAK. Therefore HD: DB: : EC or DB : EK; whence (VI. 17.) HD.EK = BD. But it is easy to show that BF: GC :: HD: EK; and therefore (VI. 1.) BF.GC: GC? :: HD.EK or BD’: EKP; whence, alternately, BF.GC : BD? :: GC?: EK. It is also plain, that GC: EK :: FG : DE or BC; and, therefore (VI. 22.) GC: EK':: FGʻ: BC. Hence (V. 11.) BF.GC: BD :: FG': BC?; and, alternately, BF.GC: FGʻ :: BDP : BC.
Schol. If, as in the figure for the 47th proposition of the first book of Euclid, BD, BC be equal, the rectangle BF.GC is equal to the square of FG; and therefore (VI. 17.) FG is a mean proportional between BF and GC. In this case also, it is manifest that if AB, AC be equal, the hypotenuse is trisected in F and G.
If, again, BD be the side of a square, of which BC is the diagonal, so that the square of BC is double of the square of BD, the square of FG is double of the rectangle BF.GC.
PROP. XXVIII. THEOR. LET A, B, C be the centres of three unequal circles, and let the straight lines passing through them be cut in D, E, F by exterior common tangents to each pair : D, E, F are in the same straight line.
Join DE, EF, and draw CG parallel to FE: and for brevity
put a, b, c to denote respectively the radii from the centres A, B, c. Then, (App. I. 25.) c:a:: CF: AF. But (VI. 2.) CF: AF:: GE : AE: therefore (V. 11.) c:a:: GE : AE. Also, (App. I.25.) a:1::AE: BE; therefore, ex æquo, c:6::GE: BE.
Again, c:b:: CD: BD : therefore (V. 11.) GE:BE:: CD: Bð; whence (VI. 2.) GC, DE are parallel
. Hence, since (const.) EF is parallel to GC, DE, EF are in the same straight line ; for if they were not, two intersecting straight lines would be both parallel to the same straight line, which (I. 29. cor. 2.) is impossible.
PROP. XXIX. THEOR.
IF through the point of bisection of the base of a triangle, a straight line be drawn cutting one of the other sides and the remaining one produced, and cutting a parallel to the base through the vertex; the straight line intercepted between the two extreme points thus obtained, is divided harmonically by the other two.
Let ABC be a triangle, D the middle point of its base, and AE a parallel to CB; draw any straight line through D, cutting AB, AE, CA produced, in F, E, G : then DG : GE:: DF: FE.
For, in the similar triangles GCD, GAE, GD: GE :: CD or BD : EA: and in the similar triangles BDF, AEF, BD : EA : DF: FE. Hence (V. 11.) GD: GE :: DF: FE: and therefore (VI. def. 5.) DG is divided harmonically in F and E.
In the same manner it would be shown that G'D: G'E'::DF: FE'; and therefore E'G' is divided harmonically in D and F.
Cor. Hence, if through D the middle point of the base, DFG be drawn, and if AE be drawn, so that GD is divided harmonically in E and F, AE is parallel to BC. For, if not, draw AH parallel to BC. Then, by this proposition, GH: HF::GD : DF: and therefore (V. 11.) GE: EF :: GH: HF, which (V. 14.) is absurd. Therefore AE is parallel to BC.
PROP. XXX. THEOR.
IF through the extremities and points of section of a straight line divided harmonically, straight lines be drawn to any point without the line, these four lines divide harmonically any straight line that falls upon them, or their continuations.
Let AB be a straight line divided harmonically in C and D, and E a point without it; join EA, EC, ED, EB: any straight line FGHK, falling on these, is divided harmonically in G and H.
Through D draw LM (APP. I. 7.
2.) so that LD may be equal to DM, and through H draw NO parallel to LM. Then, in the similar triangles LDE, NHE, and MDE, OHE, LD : NH:: DE: HE, and DM : HO::DE: HE, whence (V. 11.) LD: NH :: DM: HO; and therefore (V. 14.) because LD, DM are equal, NH is equal to HO. Again, since LD, DM are equal, EF (APP. I. 29. cor.) is parallel to LM, and therefore (1. 30.) to NO: and since FGHK is drawn through the middle point of NO, the base of the triangle ENO, it is divided harmoni. cally (App. I. 29.) in G and H.
PROP. XXXI. THEOR.
If through any point in the diameter of a circle produced, two straight lines be drawn making equal angles with the diameter, and cutting the circumference in four points, the straight lines joining the opposite points intersect one another in the diameter: and their point of intersection and the circumference divide the produced diameter harmonically.
Let A be a point in the diameter BC produced, and let the straight lines ADE, AFG, making equal angles with AB, cut the
circumference in D, E, F, G: the straight lines joining DG, EF, cut BC in the same point H, and AB is divided harmonically in H and C.
Join BE, BG, DB, DC. Then (III. 8.) AD is equal to AF, and AE to EG; and the angles ADG, AFE are equal, being supplements of the angles EDG, EFG in the same segment. Hence the triangles in which are these two angles and the equal angles at A, and which have the equal sides AD, AF, have (I. 26.) their sides which are op- 3 posite to ADG, AFE, equal, and consequently DG, FE must cut AB in the same point H. Now, since AE, AG are equal, AB common, and the angles at A equal, the chords BE, BG are (I. 4.) equal, and therefore (III. 28.) the arcs themselves. Hence (III. 27.) the angles BDE, BDG are equal, and BD bisects the exterior angle EDH of the triangle ADH. Again, BDC is (III. 31.) a right angle, and is therefore equal to the two BDE, ADC: from these equals take the equals BDG, BDE, and the remaining angles CĐH, CDA are equal : and therefore, since DC, DB bisect the interior and exterior angles at the vertex of the triangle ADH, BA is divided harmonically (VI. A. cor.) in C and H.
Schol. 1. Since BA : AC:: BH : HC, the ratio of BH to HC is always the same, so long as A is a fixed point: and therefore, however the positions of the points D, E, F, G may vary, provided the angles at A are equal, the transverse lines DG, FE will always cut the diameter in the same point H.
Schol. 2. Conversely, were a point H taken in the diameter, and transverse chords DG, EF drawn through it making equal angles with the diameter, it would be shown in a similar manner, that the straight lines EDA, GFA, each passing through the extremities that lie on the same side of the diameter, would always meet in the same point in the continuation of the diameter, and that BA: CA :: BH : HC.
PROP. XXXII. THEOR.
IF in the diameter of a circle and its continuation, two points be taken on the opposite sides of the centre, such that the rectangle under their distances from the centre may be equal to the square of the radius, any circle whatever described through these points, bisects the circumference of the other circle.
Let ABC be a circle, D a point in any diameter, and E a point
in the same diameter produced on the other side of the centre F: then, if the rectangle DF.FE be equal to the square of the radius, any circle GDHE described through D, E, bisects the circumference of ABC.
Through G, one of the points of intersection of the circles, draw GFH a chord of GHE. Then (III. 35.) the rectangle GF.FH is equal to the rectangle DF.FE, that is, (hyp.) to the square of the radius GF. Hence GF, FH are equal, and H is a point in the circumference of the circle ABC. But GH is drawn through its centre, and is therefore a diameter; wherefore its circumference is bisected in the points G, H.
PROP. XXXIII. THEOR.
A straight line drawn from the point of intersection of two tangents of a circle to the concave circumference, is divided harmonically by the convex circumference and the chord joining the points of contact.
Through the point A let the tangents AB, AC be drawn to the circle BFC, and let BC join the points of contact; any straight line ADEF drawn from A to the concave circumference, is divided harmonically in D and E.
For (III. 17. schol.) AB, AC are equal ; and therefore (II. 5. cor. 5.) AB2 or (III. 36.) FA.AD = AE+ BE.EC; or (II. 1. and III. 35.) FE.AD + EA.AD =AE + FE.ED= II. 2.) AE.ED + EA.AD + FE.ED. Take away EA.AD; then, since (II. 1.) AE.ED + FE.ED = AF.ED, the remainder becomes FE.AD = AF.ED; whence (VI. 16.) AF: AD : : EF: ED.
If four circles be described, either all outside, or all inside, of any quadrilateral, each of them touching three of its sides; the circumference of a circle will pass through all their centres.
Let ABCD be the quadrilateral, and E, F, G, H the centres of the external circles, each touching three sides ; a circle may be described through E, F, G, H.
Join AE, BE, BF, &c. Then, since AE, AH bisect two vertical opposite angles, HAD is equal to EAK, and (I. 14. cor.)