AC, BD will be equal to the fum of the rectangles under the oppofite fides, viz. to the rectangle under AB and DC, together with the rectangle under AD and BC.

A

ང་

B

C

E

D

For make the angle A B E equal to the angle e B D. Then the triangles A BD, BCE will be fimilar, because the angles BDA, BCA being both in the fame fegment, are [by 21.3.] equal to one another, and the angle A B D is equal to the angle EBC, because of the : addition of the common angle BEBD. to the equal ones A BE, DBC. Therefore as B C is to OC E, fo will BD be to DA; and fo [by 16. 6.] the rectangle under BC and D A, will be equal to the rectangle under CE and BD. Alfo the triangles A B E, CBD will be fimilar, because the angles BAE, BDC in the fame segment are [by 21. 3.] equal to one another. Therefore as A E is to CAB, fo will, CD be to BD. Confequently the rectangle under AB and CD will [by 16. 6.] be equal to the rectanIgle under A E and B D. But the rectangle under BD and Ac [by 1. 2.] is equal to the rectangle under A E and BD, and that under EC and BD. Therefore fince it has been proved, that the rectangle under CE and BD is equal to the rectangle under BC and DA, and the rectangle under AE and BD equal to the rectangle under A B and CD; the rectangle under AC and BD will be equal to the fum of the rectangles under BC and A D, and A B and DC.

Therefore, &c. Which was to be demonstrated.

This is a moft elegant and ufeful propofition, being the fountain and foundation of many fine theorems and ufeful efflections, relating to the properties of right lines drawn in a circle. The theorem is Ptolemy's, and is to be found -in his Almagest.

PROP. XIV. THE OR. If two tangents CA, CB to a circle be drawn from any point c, and from any right line d be drawn cutting the circumference of the circle in the points E, Dand the trapezium AEBD be com

pleted

t

A

pleted within the circle; the rectangles under the oppofite fides AE and D B, and AD and E B will be equal to one another.

C

E

For because [by 32. 3] the angles CA E, CBE made by the right lines AE, EB are equal to the angles A D E. ADB in the alternate fegments of the circle, B the triangles ACE, DAC; and

BCE, BCD will be fimilar : therefore it will be [by 4. 6.] as AE is to AC, fo is AD to DC ; \ and as DB to DC, fo is BE to CB. Therefore [by prop. 1. of the additions to the fixth book] the four rectangles under A E and DB; AC and DC; AD and BE; and DC and CB, will be proportional. But because the tangents AC, BC are equal, and fo the rectangle under AC and DC equal to the rectangle under Cв and DC. Therefore [by 9. 5.] the rectangle under AE and DB will be equal to the rectangle under AD and B É.

Therefore, &c. Which was to be demonftrated.

PROP. XV. THEOR.

In a circle if AB be the diameter, and any right line CD cuts the fame in the point L, and from the extremes A and B of that diameter two perpendiculars AE, BF be drawn to that right line CD; the fegments CF, ED of it will be equal to one another.

For join the points E, B, and from the centre 1 draw IG perpendicular to CD, and continue it to H in the line E B.

Then [by 3.3.] will CG be equal to GD. And because IG, A E are parallel, and fince BI is equal to IA; BH will be [by 4. 6.] equal to HE, and FG to GE. Therefore the difference between CG and FG

Will be equal to the difference between GD and GE, that is CF will be equal to ED.

Therefore, &c. Which was to be demonftrated.

Corollary. Hence if there be any trapezium whofe two oppofite angles are right angles, and a diagonal be drawn joining these right angles; and if from each of the other angles be drawn a perpendicular upon that diagonal; the fegments of this diagonal made by those perpendiculars will be equal to one another.

PRO P. XVI.

If CD, BD be two tangents to a circle at B and c, whereof c is the extreme of a diameter AC, and the perpendicular вE be drawn upon a c, and if the points A and D be joined by a right line cutting the perpendicular в E in the point F; this perpendicular B E will be bifected in the point r.

For thro' B draw A B to meet the tangent C D (continued out) in the point G; and join в C.

Then because the tangents BD, DC are equal, the angles DCB, DBC [by 5. 1.] will be equal, and the angles DBC, DBG are equal to a right angle, because the angle CBG, which is equal to the angle ABC, is [by 31. 3.]a right angle. Also the angles DCB, GGB [by 32. 1.] are equal to a right angle. Wherefore the angle DBG will be equal to the angle

G

B

DGB. And fo [by 6. 1.] BD,

that is CD, will be equal to DG. Wherefore fince BE, GC are par-A allel, EF will be equal to F B.

F

E с

Therefore, &c. Which was to be demonftrated.

Corollary. Hence the perimeter or circuit of any regular polygon circumfcribing a circle, is to the perimeter of a regular polygon of half the number of fides inscribed in that circle, as the diameter CA is to the fegment AE thereof.

PROP. XVII. THEOR.

If the point в be taken in the diameter co of a circle, whofe centre is E, and the point A in the continuation of that diameter fuch, that the difference between AC and CB be to CB, as CB is to BE: two right lines AF, BF drawn from thofe affumed points A and в to any point F in the circumference of the circle, will be in a conftant ratio,

1

viz. as A c is to C B.

For make A D equal to the difference between A C and CB, and draw the femidiameter EF.

F

A D C

B E

it will be [by 12.

Then because [by conftruction] AD is to DC, as CB is to BE; that is, [by compounding by the G 18th 5.] AC is to CD, or CB, as CE is to EB: 5.] as A E is to CE, that is, EF, fo is CE or E F to E B. Therefore because the triangles A EF, FEB have one angle E common to both, and the fides about it proportional: alfo the remaining fides AF, FB [by 6.6] will be to one another as A E is to E F, or E C, that is, CE to E B. But as CE is to E B, fo is AC to C B. Wherefore it will be as A F is to F B, fo is AC to C B. Therefore, &c. Which was to be demonftrated.. This is a famous and useful propofition of Apollonius's, SCHOLI U M.

By means of this propofition it is easy to find a point N within a triangle ABC fuch that drawing to it three right lines from the three angles of that triangle, they shall be in given ratios, viz. respectively as the lines R, S, T.

For divide the fide AB of the given triangle in the point Dfo, that AD be to DB, as R is to s, (which may be cafily done by putting the lines, R, s in the fame direction, and then [by 10. 6.] dividing the line AB in the fame manner as the fum of the lines R and s is divided.) This done, find [by 12. 6.] a fourth proportional right line B E to AB, DB and the difference between AD and DB. Bifect DE in F, and about F with the distance DF or F E defcribe a circle. Again, divide the fide AC of the given triangle fo in 1, that

ting the former one in the point N.

eight lines

AN, BN, CN, and thefe will be to one another in the given ratios of R, S, T.

Or the centre F may be found by taking BF a third proportional to DB and the difference between AD and DB. And So may the centre G be found, by taking CG equal to a third proportional between IC and the difference between AI and

I C.

THEOR.

PRO P. XVIII. In a circle the ratio of a greater arch BG to a leffer A B, will be greater than that of the right line BG joining the ends of the greater arch, to that of the right line A B joining the ends of the leffer arch a B. For draw the right line AG, and the equal right lines AD, DG, and join the points D, B by a right line cutting AG in E, and draw the perpendicular pz to AG and having described the arch HTV about D, as a centre, with the distance D E, continue out Dz to meet the fame in T. Now because the line DB

bifects the angle ABG, for the lines A D, DG being equal [by conftruction] the arches AD, GD and fo the angles A B D, DBG [by 29. and 21. 3. will be equal. Therefore [by 3. 6.] as в G is to A B, fo will EG be to É A. And fince BG is greater than AB; EG will be greater than A E.

B

T

G

HE Z

Where

D

X 2

fore