DCE have one angle at A equal to one at D, and the sides about these angles proportionals, viz. BA to AC, as CD to DE, the triangle ABC is equiangular (6. 6.) to DCE: therefore the angle ABC is equal to the angle DCE; and the angle BAC was proved to be equal to ACD therefore the whole angle ACE is equal to the two angles ABC, BAC; add the common angle ACB, then the angles ACE, ACB are equal to the angles ABC, BAC, ACB but ABC, BAC, ACB are equal to two right angles (32. 1.); therefore also the angles ACE, ACB are equal to two right angles and since at the point C, in the straight line AC, the two straight lines BC, CE, which are on the opposite sides of it, make the adjacent angles ACE, ACB equal to two right angles; therefore (14. 1.) BC and CE are in a straight line. Wherefore, if two triangles, &c. Q. E. D.

PROP. XXXIII. THEOR.

In equal circles, angles, whether at the centres or circumferences, have the same ratio which the circumferences on which they stand have to one another: so also have the sectors.*

Let ABC, DEF be equal circles and at their centres the angles BGC, EHF, and the angles BAC, EDF at their circumferences as the circumference BC to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF; and also the sector BGC to the sector EHF.

Take any number of circumferences CK, KL, each equal to BC, and any number whatever FM, MN, each equal to EF: and join GK, GL, HM, HN. Because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also all equal (27. 3.): therefore what multiple soever the circumference BL is of the circumference BC, the same multi ple is the angle BGL of the angle BGC: for the same reason, whatever multiple of the circumference EN is of the circumference EF, the same multiple is the angle EHN of the angle EHF and if the circumference BL be equal to the circumfer

See Notc.

ence EN, the angle BGL is also equal (27. 3.) to the angle EHN; and if the circumference BL be greater than EN, likewise the angle BGL is greater than EHN; and if less, less: there being then four magnitudes, the two circumferences BC, EF, and the two angles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equimultiples whatever, viz. the circumference BI., and the angle BGL; and of the circumference EF, and of the angle EHF, any equimultiples

whatever, viz. the circumference EN, and the angle EHN: and it has been proved, that if the circumference BL be greater than EN, the angle BGL is greater than EHN: and if equal, equal; and if less, less: as therefore the circumference BC to the circumference EF, so (5. def. 5.) is the angle BGC to the angle EHF: but as the angle BGC is to the angle EHF, so is (15. 5.) the angle BAC to the angle EDF, for each is double of each (20. 3.): therefore, as the circumference BC is to EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF.

Also, as the circumference BC to EF, so is the sector BGC to the sector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, Ó, and join BX, XC, CO, OK: then, because in the triangles GBC, GCK, the two sides BG, GC are equal to the two CG, GK, and that they contain equal angles; the base BC is equal (4. 1.) to the base CK, and the triangle GBC to the triangle GCK: and because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference of the circle ABC, is equal to the remaining part of the whole circumference of the same circle: wherefore the angle BXC is equal to the angle COK (27. 3.), and the segment BXC is therefore similar to the segment COK (11. def.3.):

and they are upon equal straight lines BC, CK: but similar segments of circles upon equal straight lines, are equal (24. 3.) to one another: therefore the segment BXC is equal to the segment COK: and the triangle BGC is equal to the triangle CGK; therefore the whole, the sector BGC, is equal to the whole, the sector CGK for the same reason, the sector KGL is equal to each of the sectors BGC, CGK in the same manner, the sectors EHF, FHM, MHN may be proved equal to one another: therefore, what multiple soever the circumference BL is of the circumference BC, the same multiple is the sector BGL of the sector BGC: for the same reason, whatever multiple the circumference EN is of EF, the same multiple is the sector EHN of the sector EHF and if the circumference BL be equal to

EN, the sector BGL is equal to the sector EHN and if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if less, less: since, then, there are four magnitudes, the two circumferences BC, EF, and the two sectors BGC, EHF, and of the circumference BC, and sector BGC, the circumference BL and sector BGL are any equal multiples whatever: and of the circumference EF, and sector EHF, the circumference EN and sector EHN are any equimultiples whatever; and that it has been proved, if the circumference BL be greater than EN, the sector BGL is greater than the sector EHN; and if equal, equal; and if less, less. Therefore (5. def. 5), as the circumference BC is to the circumference EF, so is the sector BGC to the sector EHF. Wherefore, in equal circles, &c. Q. E. D.

PROP. B. THEOR.

Ir an angle of a triangle be bisected by a straight line, which likewise cuts the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.*

Let ABC be a triangle, and let the angle BAC be bisected by the straight line AD; the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD.

Α

Describe the circle (5. 4.) ACB about the triangle, and produce AD to the circumference in E, and join EC: then because the angle BAD is equal to the angle CAE, and the angle ABD to the angle (21. B 3.) AEC, for they are in the same segment: the triangles ABD, AEC are equiangular to one another: therefore as BA to AD, so is (4. 6.) EA to AC, and consequently the rectangle BA, AC is equal (16. 6.) to the rectangle EA, AD, that is

:

C

Ꭰ

E

(3. 2.), to the rectangle ED, DA, together with the square of AD but the rectangle ED, DA is equal to the rectangle (35. 3.) BD, DC. Therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, &c. Q. E. D.

PROP. C. THEOR.

Ir from any angle of a triangle a straight line be drawn perpendicular to the base; the rectangle contained by the sides of the triangle is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.*

Let ABC be a triangle, and AD the perpendicular from the angle A to the base BC; the rectangle BA, AC is equal to the rectangle contained by AD and the diameter of the circle described about the triangle.

* See Notes.

Describe (5. 4.) the circle ACB about the triangle, and draw its diameter AE, and join EC: because the right angle BDA is equal (31. 3.) to the angle ECA in a semicircle, and B the angle ABD to the angle AEC in the same segment (21. 3.); the triangles ABD, AEC are equiangular: therefore as (4. 6.) BA to AD, so is EA to AC; and consequently the rectangle BA, AC is equal (16. 6.) to

the rectangle EA, AD. If, therefore, from an angle, &c. Q. E. D.

PROP. D. THEOR.

THE rectangle contained by the diagonals of a quadrilateral inscribed in a circle, is equal to both the rectangles contained by its opposite sides.*

Let ABCD be any quadrilateral inscribed in a circle, and join AC, BD; the rectangle contained by AC, BD is equal to the two rectangles contained by AB, CD, and by AD, BC.t

Make the angle ABE equal to the angle DBC; add to each of these the common angle EBD, then the angle ABD is equal to the angle EBC; and the angle BDA is equal (21. 3.) to the angle BCE, because they are in the same segment; therefore the triangle ABD is equiangular to the

triangle BCE; wherefore (4. 6.) as BC is to CE, so is BD to DA; and consequently the rectangle BC, AD is equal (16. 6.) to the rectangle BD, CE: again, because the angle ABE is equal to the angle DBC, and the angle (21. 3.) BAE to the angle BDC, the triangle ABE is equiangular to the triangle BCD: as therefore BA to AE, so is BD to DC; wherefore the rectangle BA, DC is equal to the

B

E

D

A

rectangle BD, AE: but the rectangle BC, AD has been shown equal to the rectangle BD, CE; therefore the whole rectangle AC, BD (1. 2.) is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefore the rectangle, &c. Q. E. D.

See Note.

†This is a Lemma of Cl. Ptolomæus, in page 9 of his μyaan σurağı.

Bb