A let AB be parallel to DC, and AC to DE: BC and CE are in a straight line. Because AB is parallel to DC, and the straight line AC meets them, the alternate angles BAC, ACD are equal (29. 1.): For the same reason, the angle CDE is equal to the angle ACD; wherefore also BAC is equal to CDE: And because the triangles ABC, 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. vi.) to BCE; 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, etc. Q. E, D. PROPOSITION XXXIII. THEOR. In equal circles, angles, whether at the centres or circumfer ences, 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. 111.); therefore what multiple soever the circumference BL is of the circumference BC, the same multiple is the angle BGL of the angle BGC: For the same reason, whatever multiple 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 circumference EN, the angle BGL is also equal (27. III.) to the angle EHN; and if the circumference BLA A 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 ngles BGC, EHF; of the circumference BC, and of the angle BGC, have been taken any equi N K F P. multiples whatever, viz. the circumference BL, 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. v.) is the angle BGC to the angle EHF: But as the angle BGC is to the angle EHF, so is (15. v.) the angle BAC to the angle EDF, for each is double of each (20. 111.); 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 také any points X, O, and join BÝ, 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: therefore the angle BXC is equal (27. 111.) to the angle COK; and the segment BXC is therefore similar to the segment COK (11 Def. 111.); and they are upon equal straight lines BC, CK: But similar segments of circles upon equal straight lines are equal (24. 111.) 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 cir. cumference 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 equimultiples 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. v.), as the circumference BC is to the circumference EF, so is the sector BGC to the sector EHF. Wherefore, in equal circles, etc. Q. E. D. H N M PROPOSITION B. THEOR. If 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. Describe the circle (5. iv.) 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. 11.) 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. vi.) EA to AC; and consequently the rectangle BÀ, AC is equal (16. vi.) to the rectangle EA, AD; that is (3. 11.), to the rectangle ED, DĂ, together with the square of AD: But the rectangle ED, DA is equal to the rectangle (35. III.) BD, DC; therefore the rectangle BA, AC is equal to the rectangle BD, DC, together with the square of AD. Wherefore, if an angle, etc. Q. E, D. PROPOSITION C. THEOR. If from any angle of a triangle a straight line be drawn per pendicular 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. Describe (5. iv.) the circle ACB about the triangle, and draw its diameter AE, and join EC: Because the right angle BDA is equal (31. III.) to the angle ECA in a semicircle, and the angle ABD to the angle AEC in the same segment (21. 11.); the triangles ABD, AEC are equiangular: Therefore as (4. vI.) BA to AD, so is EA to AC; and consequently the rectangle BA, AC is equal (16. vi.) to the rectangle EA, AD. If, therefore, from any angle, etc. Q. E. D. PROPOSITION 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.* * This is a Lemma of Cl. Ptolomæus, in page 9 of his Msgáan Zúrvaltısı 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. 11.) to the angle BCE, because they are in the B same segment; therefore the triangle ABD is equiangular to the triangle BCE: wherefore (4. vi.), as BC is to CE, so is BD to DA; and consequently the rectangle BC, AD is equal (16. vi.) to the rectangle BD, ČE: Again, because the angle ABE is equal to the angle DBC, and the angle (21. III.) 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 rectangle BD, AE: But the rectangle BC, AD has been shown equal to the rectangle BD, CE; therefore the whole rectangle AC, BD (1. 11.) is equal to the rectangle AB, DC, together with the rectangle AD, BC. Therefore, the rectangle, etc. Q. E, D. EXERCISES ON Book VI. 1. If perpendiculars be drawn from the angles of a triangle to the opposite sides : (a) They will be to one another in the reciprocal ratio of the sides upon which they fall : (6) Each pair will cut off a pair of similar right-angled triangles from the original triangle; which are to be specifically assigned, and the statement with respect to them proved. 2. Triangles which have equal or supplementary angles are to one another in a ratio compounded of the ratios of their containing sides. And conversely, if two triangles be to one another in a ratio compounded of the ratios of their sides, the angles contained by those sides are either equal or supplementary. Give, in connexion with this, the enunciation of vi. 23, in its most general form. 3. Prove . 35, 36, 37, by means of similar triangles. 4. Given the ratio of two lines, and either their sum, difference, rectangle, sum of squares, or difference of the squares, to find those lines. 5. Homologous lines will divide similar rectilineal figures into figures which are similar each to its homologous one. 6. Lines drawn homologously (whether from the angular points or otherwise) to homologous points of similar figures, will make the homologous angles equal, each to each ; and will have, each to each, the same ratio. 7. If two similar figures have one side parallel to its homologous one, then lines joining the homologous angles intersect in the same point. [In certain cases, two such points may exist. It will be a good exercise to discover those cases.] 8. Similar rectilineal figures are to one another in the ratio compounded of the ratios of any two lines anyhow drawn in the one to two lines drawn homologously to them in the other. What does this become when the homologous lines are homologous sides of the figure? And what when both are repetitions of the same side of each figure? 9. Triangles and parallelograms are to one another in a ratio compounded of the ratios of their bases and the ratio of their perpendiculars. Generalise this by taking lines equally inclined to the bases, instead of the perpendiculars. 10. Draw a perpendicular from the right angle to the hypothenuse of a right-angled triangle: then the sides about the right angle are to one another in the sub-duplicate ratio of the adjacent segments of the hypothenuse; and the diameters of the circles inscribed in the two partial triangles are to one another as the adjacent sides of the original triangle. 11. Circles inscribed, escribed, and circumscribed to similar tri. angles, have their respective diameters in the same ratio that the homologous sides of the triangles have.* 12. Having two lines given, it is required to add to them, to take from them, or to add to one and take from the other, lines in a given ratio, so that the sum, difference, or ratio of the two final lines shall be also given. 13. Through a given point to draw a line terminated by a given circle, such that the sum, difference, or ratio of its segments made at that point shall be given. 14. Given the ratio of the sides and the vertical angle of a triangle, to construct' it when the third datum is (a) The base; (h) The radius of the circle escribed to the base. 15. Given any three of the radii of the five circles, viz., the circumscribed, the inscribed, and the three escribed, to construct the triangle, [Many elegant properties are connected with the solution of this problem.] 16. Similar polygons, whether inscribed in or described about circles, have their perimeters in the ratio of the diameters of those circles, and their areas in the duplicate ratio of those diameters; and one inscribed polygon will be to the other in the same ratio as the corresponding circumscribed polygon is to the other. 17. Let two triangles ABC, A'B'C' have their bases AB, A'B' situated in the same indefinite line; and let CC' be drawn to cut this line in D: then will the triangles have to one another a ratio compounded of the ratios of the bases AB: A'B', and of the distances of their vertices from the point of intersection DC: DC'. * A circle which touches one side of a triangle and the other two sides produced is called an escribed circle. |