Let ABC, DCE be two triangles which have the two sides BA, AC proportional to the two CD, DE, viz. BA to AC as CD to DE, and let AB be parallel to DC, and AC to DE: then BC and CE shall be in a straight line. For, because AB is parallel to DC, and AC meets them, the alternate angles BAC, ACD are equal: but, for the like reason, the angle ACD is equal to the angle CDE; therefore also the angle BAC is equal to the angle CDE: And because Λ the triangles ABC, DCE have the angle at A equal to the angle at D, and the sides about the equal angles proportionals, therefore (6. 6) the triangle ABC is equiangular to the triangle DCE, and the angle ABC is equal to the angle DCE: But the angle BAC was proved equal to the angle ACD; therefore the angles ABC, BAC are together equal to the angle ACE: To each of these equals add the angle ACB; then the angles ABC, BAC, ACB are together equal to the angles ACB, ACE: But ABC, BAC, ACB are together equal to two right angles; therefore also the angles ACB, ACE are together equal to two right angles, and therefore BC and CE are in a straight line. Wherefore, If two triangles &c. PROP. XXXIII. Q. E.D. 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 and so also have the sectors. Let ABC, DEF be equal circles, and let BGC, EHF be angles at their centres, and BAC, EDF angles at their circumferences: as the circumference BC to the circumference EF, so shall the angle BGC be to the angle EHF, and the angle BAC to the angle EDF; and so also shall the sector BGC be 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: Then, because the circumferences BC, CK, KL are all equal, the angles BGC, CGK, KGL are also equal; and therefore, whatever multiple the circumference BL is of the circumference BC, the same is the angle BGL of the angle BGC: In like manner, whatever multiple the circumference EN is of the circumference EF, the same is the angle EHN of the angle EHF: Now, if the circumference BL be equal to the circumference EN, the angle BGL is also D L G H N K M B C E F equal to the angle EHN, and if greater, greater, and if less, less: But the circumference BL and the angle BGL are any equimultiples of the circumference BC and angle BGC, and the circumference EN and the angle EHN are any equimultiples of the circumference EF and the angle EHF; therefore (5. Def. 5) as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF: But as the angle BGC is to the angle EHF, so is the angle BAC to the angle EDF (5. 15), for each is double of each; therefore, as the circumference BC is to the circumference EF, so is the angle BGC to the angle EHF, and the angle BAC to the angle EDF. Also, as the circumference BC is to the circumference EF, so shall the sector BGC be to the sector EHF. Join BC, CK, and in the circumferences BC, CK take any points X, O, and join BX, XC, CO, OK: Then, because in the triangles BGC, CGK, the two sides BG, GC are equal to the two CG, GK, each to each, and that they contain equal angles-therefore the base BC is equal to the base CK, and the triangle BGC to the triangle CGK: And, because the circumference BC is equal to the circumference CK, the remaining part of the whole circumference, when BC is taken from it, is equal to the remaining part of the same circumference, when CK is taken from it: Therefore (3. 27) the angle BXC is equal to the angle COK, and the segment BXC is therefore similar to the segment COK: But they are upon equal straight lines BC, CK; 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: In like manner, the sector KGL is equal to each of the sectors BGC, CGK, and so also the sectors EHF, FHM, MHN may be proved equal to one another: Therefore, whatever multiple the circumference BL is of the circumference BC, the same is the sector BGL of the sector BGC; and whatever multiple the circumference EN is of the circumference EF, the same is the sector EHN of the sector EHF: Now, if the circumference BL be equal to the circumference EN, the sector BGL will be equal to the sector EHN, and if greater, greater, and if less, less: But the circumference BL and the sector BGL are any equimultiples of the circumference BC and the sector BGC, and the circum A D M E F N ference EN and the sector EHN are any equimultiples of the circumference EF and the sector EHF; therefore, 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. If the vertical angle of a triangle be bisected by a straight line which likewise cuts the base, the rectangle contained by the sides of the triangle shall be 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, which cuts the base BC in D: the rectangle BA, AC shall be equal to the rectangle BD, DC, together with the square of AD. About the triangle describe the circle ABC, 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 AEC, (being angles in the same segment of a circle,) the triangles ABD, AEC are equiangular to one another: Therefore BA is to AD as EA to AC; and therefore (6. 16) the rectangle BA, AC is equal to the B rectangle EA, AD, that is (2. 3) to the rectangle ED, DA, together with the square of AD: But the rectangle D A K ED, DA is equal to the rectangle BD, DC (3. 35); therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, If the vertical angle &c. Q. E.D. PROP. C. THEOR. If 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 let AD be the perpendicular from the angle A to the base BC: the rectangle BA, AC shall be equal to the rectangle contained by AD and the diameter of the circle described about the triangle. About the triangle describe the circle ABC, and draw the diameter AE, and join EC: Then, because the right angle BDA is equal to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same segment, therefore the triangles ABD, AEC are equiangular : Therefore BA is to AD as EA to AC, B and therefore the rectangle BA, AC is equal to the rectangle EA, AD. Wherefore, If from any angle &c. Q. E.D. PROP. D. THEOR. The rectangle, contained by the diagonals of a quadrilateral inscribed in a circle, is equal to the sum of 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. |