and let AC, BD be drawn; the rectangle AC.BD is equal Book VI. to the two rectangles AB.CD, and AD.BC.

Make the angle ABE equal to the angle DBC; add to each

B

of these the common angle EBD, then the angle ABD is equal to the angle EBC: And the angle BDA is equal a to a 21. 3, the angle BCE, because they are in the fame segment; therefore the triangle ABD is equiangular to the triangle BCE. Whereforeb, BC:CE:: BD: DA, and confequently c BC.DA BD.CE. Again, because the angle ABE is equal to the angle DBC, and the angle a BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD; therefore BA: AE:: BD: DC, and BA.DCBD.AE; But it was fhewn that BC.DA

E

D

b 4. 6.

c 16. 6.

C

BD.CE; wherefore BC.DA+BA.DCBD.CE+BD.AE= BD.AC. That is the rectangle contained by BD and AD, d1.2. is equal to the rectangles contained by AB, CD, and AD, BC. Therefore the rectangle, &c. Q. E. D.

PROP. E. THEOR.

IF

F an arch of a circle be bifected, and from the extremities of the arch, and from the point of bisection, straight lines be drawn to any point in the circumference, the fum of the two lines drawn from the extremities of the arch will have to the line. drawn from the point of bifection, the fame ratio which the ftraight line fubtending the arch, has to the ftraight line fubtending half the arch.

Book VI.

Let ABD be a circle, of which AB is an arch, bifected in C, and from A, C and B to D, any point whatever in the circumference, let AD, CD, BD be drawn; the fum of the two lines AD and DB has to DC the fame ratio that BA has to AC.

For fince ACBD is a quadrilateral infcribed in a circle, of which the diagonals are AB a D. 6. and CD, AD.ČB+DB.AC a=

b 1. 2.

AB.CD: but AD.CB+DB.ACAD.AC+DB.AC, because CB AC. Therefore AD.AC+DB.AC, that is b, AD+DB.AC=AB.CD. And because the fides of equal € 14. 6. rectangles are reciprocally proportional, AD+DB: DĈ :: AB: AC. Wherefore, &c. Q. E. D.

PROP. F. THEO R.

IF
F two points be taken in the diameter of a circle,
fuch that the rectangle contained by the feg-
ments intercepted between them and the centre of
the circle be equal to the fquare of the radius; and
if from these points two ftraight lines be drawn
to any point whatsoever in the circumference of the
circle, the ratio of thefe lines will be the fame with
the ratio of the fegments intercepted between the
two firft-mentioned points, and the circumference of
the circle.

Let ABC be a circle, of which the centre is D, and in DA produced, let the points E and F be fuch that the rectangle

ED,

ED, DF is equal to the fquare of AD; from E and F to any Book VI.
point B in the circumference, let EB, FB be drawn;
FB: BE:: FA: AE.

Join BD, and because the rectangle FD, DE is equal to the fquare of AD, that is, of DB, FÐ : DB : : DB : DE a. The

a 17. 6.

6. 6.

two triangles, FDB, BDE have therefore the fides propor-
tional that are about the common angle D; therefore they
are equiangular b, the angle DEB being equal to the angle
DBF, and DBE to DFB. Now, fince the fides about these
equal angles are alfo proportional c, FB: BD:: BE: ED, and c 4. 6.
alternately 4, FB: BE:BD: ED, or FB: BE:: AD: DE. d 16.5.
But because FD: DA:: DA: DE, by divifione, FA: DA::
AE: ED, and alternately e, FA: AE::DA: ED. Now, e 17. 5.
it has been fhewn that FB: BE::AD: DE, therefore
FB: BE FA: AE. Therefore, &c. Q. E. D.

fit. S.

3.6.

COR. If AB be drawn, because FB: BE:: FA: AE, the angle FBE is bifected g by AB. Alfo, fince FD: DC:: DC: DE, by compofition h, FC: DC::CE: ED, and fince it h18. 5. has been fhewn that FA: AD (DC) :: AE: ED, therefore, ex æquo, FA: AE:: FC: CE: But FB: BE: FA: AE, therefore, FB: BE:: FC: CE f; fo that if FB be produced to G, and if BC be drawn, the angle EBG is bifected by the line BC k;

k A. 6.

1

PROP.

Book VI:

PROP. G.

THEOR.

IF from the extremity of the diameter of a circle a ftraight line be drawn in the circle, and if either within the circle or produced without it, it meet a line perpendicular to the fame diameter, the rectangle contained by the ftraight line drawn in the circle, and the fegment of it, intercepted between the extremity of the diameter and the perpendicular, is equal to the rectangle contained by the diameter, and the fegment of it cut off by the perpendicular.

Let ABC be a circle, of which AC is a diameter, let DE be perpendicular to the diameter AC, and let AB meet DE in F; the rectangle BA.AF is equal to the rectangle CA.AD. Join BC, and because ABC is an angle in a femicircle, it is a

a 31. 3. right angle a: Now, the angle ADF is also a right angle b; b Hyp. and the angle BAC is either the fame with DAF, or vertical

to it; therefore the triangles ABC, ADF are equiangular, and c4. 6. BA: AC:: AD: AFc; therefore alfo the rectangle BA.AF,

contained by the extremes, is equal to the rectangle AC.AD d 16.6. contained by the means. d. If therefore, &c. Q. E. D.

PROP.

Book VI.

TH

PROP. H. THEOR.

HE perpendiculars drawn from the three angles of any triangle to the oppofite fides interfect one another in the fame point.

Let ABC be a triangle, BD and CE two perpendiculars interfecting one another in F; let AF be joined, and produced if neceffary, let it meet BC in G, AG is perpendicular to BC.

Join DE, and about the triangle AEF let a circle be defcribed, AEF; then, because AEF is a right angle, the circle defcribed about the triangle AEF will have AF for its diameter a. In the fame manner, the circle described about the triangle ADF has AF for its diameter; therefore the points A, E, F and D are in the circumference of the fame circle. But because the angle EFB is equal to the angle DFC b, and alfo the angle BEF to the angle CDF, being both right B

angles, the triangles BEF

E

a 31. 3.

b 15. I.

and CDF are equiangular, and therefore BF: EF:: CF : FDc, c 4. 6. or alternately 4, BF: FC:: EF: FD. Since, then, the fides d 6. 5. about the equal angles BFC, EFD are proportionals, the triangles BFC, EFD are alfo equiangulare; wherefore the angle e 6. 6. FCB is equal to the angle EDF. But EDF is equal to EAF, because they are angles in the fame fegment f; therefore the fat. 3. angle EAF is equal to the angle FCG: Now, the angles AFE, CFG are alfo equal, because they are vertical angles; therefore the remaining angles AEF, FGC are alfo equal: But 8 32. 1. AEF is a right angle, therefore FGC is a right angle, and AG is perpendicular to BC. Q. E. D.

COR.