LESSONS IN (I! 



88, 



by the last proposition, the triangles are equal j therefore angle 

 angle u A r, or u A bisects the angle at B A 



\XXlI.-The straight lino* drawn perpendi- 



euliir to tin- sides of a triangle through their un<l ll<> points meet 



in a point. 



t tho sides A B, BC (Fig. :i2) of the triangle ABC in 

 . and lot D o, E o perpendicular 



to A B, B c meet in G ; draw o r per- 



}>, mlieuhir to A'; then shall Y be 



the middle point of AC. Join A u, 

 , ; then Miioe A D = D B, and 



D O is common and at right angles, 



base A G base G B. Similarly, be- 

 cause B K = E c, and K G is common 



and at right angles, therefore B G = 



O c, therefore A o = o c ; and be- 



A o = a c, and G F is common, also right angle o r A = 



right angle G F c, therefore the triangles UFA, o F c are equal 



by Proposition XXX., and A F = F c. Q. E. D. 



PROPOSITION XXXIII. The straight lines which bisect one 



interior and two exterior angles of a triangle meet in a point. 



Produce A B, A c (Fig. 

 33), sides of the triangle 

 A B c, to D and F, and 

 bisect the exterior angles 

 D B c, B c F by the lines 

 B G, c o meeting in o ; 

 join o A ; then o A shall 

 bisect the angle BAC. 

 Draw G D, G E, OF pcr- 



Pig. 33. w r pendicular to A B, B c, 



c A ; then in the triangles 



O B D, G B E, because angle G B D = angle G B B by construction, 



and the right angle o D B = right angle G E B, also side G B is 



common, therefore the triangles are equal ; therefore o D = 



a E (Euc. I. 26). Similarly G E = G F, therefore Q D = OF; 



and because G D = G F, and G A ia common, also right angle 



G D A = right angle G F A, therefore, by Proposition XXX., tho 



triangles G D A, G F A are equal, and the angle GAD equals the 



angle G A F that is, G A bisects the angle BAC. Q. E. D. 

 PROPOSITION XXXIV. If two triangles have one side, and 



one angle in the one equal to one side and one angle in the 



other, and likewise their areas equal, then shall also their other 



sides and angles be equal to 



each. 



Let ABC, D c F (Fig. 34) 



be the two triangles of equal 



area, and let them be placed 



so that B c, c F, their equal 



sides, are in the same straight 



line, and B c A, c F D their 



equal angles ; then because 



C 



rig. 34. 



angles B c A, c F D are equal, A c is parallel to D F. Join A D ; 

 then because A B c, D c F are equal triangles upon equal bases 

 in the same straight line, A D is parallel to B F (Euc. I. 40) ; 

 therefore A D F c is a parallelogram. Therefore, by Euclid I. 34. 

 A c is equal to D F ; and therefore, by Euo. I. 4, the triangles arc 

 equal in every respect. Q. E. D. 



PROPOSITION XXXV. If the bases of two equal triangles 

 be in the same straight line, and tho lino joining their vertices 



bo parallel to this line, their 

 bases will bo equal. 



This is evidently the con- 

 verse of Euclid I. 40. Let ABC, 

 DBF (Fig. 35) bo two equal 

 triangles having their bases B c, 

 E F in the same straight lino, 

 and let AD, tho line joining 

 their vertices, be parallel to 

 B s 1 ; then shall B c bo equal to E p. For if B c bo not equal 

 to K F, one of them must bo tho greater. Let E F bo greater, 

 and from E F cat off E G, equal to B c, and join D o ; then 

 because ABC, D E G are two triangles upon equal bases, 

 B c, is G, in the same straight line, and between the eamo 

 parallels B G, A D, they an equal (Euc. I. 38). But by the pro- 

 position the triangle A B c is equal to the triangle DBF; there- 

 fore the triangle D E F is equal to tho triangle DEO, the greater 

 equal to the less- which is absurd. Hence E o ia not equal to 



C E 

 Pig. 35. 



O. And similarly it may be proved that ao line bt >r ia 

 qua! to B o t henoe B r B c. (J. K. D. 



PROPOSITION XXX VI. In the figure of Euclid I. 5, if A c 

 (Pig. 36) be bisected in u, and co be equal to c A, UMO BO shall 

 be equal to twice B u. 



From o draw c K parallel to A u, meeting no ia x (Koe. 

 I. 31 1, and join A K ; then the Irian* !* A x c, B K o are 

 upon the tome bate K c, and between the -nne para'Vs u r, 

 A B ; hence they are eqaal (Eoc. L 37). Again, the triaaclea 

 A K c, a K c are upon equal baeee A c, c o. and hare the MUD* 

 vertex K ; henoe they are equal (Eoc. I. 38) j but A K O IB 

 B K c ; therefore triangle B K c > triangle o x a Hence, by 

 the laet proportion, B x ~ x o ; therefore B a ia doable B x. 

 Again, became BK = xo, triangle BXC = half-triangle BOCJ 

 and became AH == HO, triangle BMC = half-triangle BAC; 

 but because A c = c u, triangle B A c = triangle B o c, and the 

 halves of equal triangle* are equal ; therefore triangle B C = 

 triangle B x c, and the angle 

 B c x is equal to tho alter- 

 nate angle ABC (Euo. 1.29), 

 and A B c is equal to A c B 

 (Euc. I. 5). Hence the 

 angle B c K is equal to the 

 angle B c u. Henoe the 

 two triangles B u c, B K c 

 have one side BC and one 

 angle Benin tho one equal 

 to one side B c and one 

 angle B c K in tho other ; 

 and their areas aro equal, becanso, by Proposition XXXV., they 

 are equal in every respect ; therefore, B u B K = haJf B o. 

 Q. E. D. 



Our next article will embrace the whole of Book I., and will 

 contain proofs of the following propositions : 



PROPOSITION XXXVII. If the diagonals of a foar-ided 

 figure bisect each other, it ia a parallelogram. 



Corollary 1. If the diagonals be equal as well as bisecting 

 each other, the figure is rectangular. 



Corollary 2. Hence tho angle in a semicircle is a right angle. 



PROPOSITION XXXVIII. If the diagonals of a four-aided 

 figure bisect each other at right angles, the figure is a rhombus. 



Corollary. If the diagonals bo also equal, it is a square. 



PROPOSITION XXXIX. If a four-sided figure have its oppo- 

 site sides equal, it is a parallelogram. 



PROPOSITION XL. If A B, BC, CD, DA be the sides of a 

 parallelogram taken in order, and points K, r, o, H be tales in 

 them such that A E = c o, and B r = D H, the figure E r o n 

 shall be a parallelogram. 



PROPOSITION XI. I. If ABC be any triangle, and DB the 

 line joining the middle points of the sides A B, A c, then sh&T 

 D E be parallel to B c, and B c shall be double D E. 



Corollary 1. The angles of the triangle DFE are equal to 

 the angles of the triangle BAC, where r is tho middU 

 of BC. 



Corollary 2. Tho sides of tho triangle DFX are each equal 

 to one-half the corresponding sides of the triangle MAC, aad 

 tho area equal to one- fourth of tao whole triangle BAC. 



PROPOSITION XL1I. If o bo any point within a triangle. 

 ABC, and D, E, tho middle points of A B, A c, be joined with r a.. 

 tho middle points of O B, o c, the figure D x o F will be a paral- 

 lelogram. 



PROPOSITION XLIIL If in the last proposition R x be the 

 middle points of o A, B c respectively, the six-sided figure 

 D K o E H shall bo equal to one-half tho triangle ABC. 



PROPOSITION XLIV. If D be the middle point of BC, the 

 side of a triangle ABC, and if A D bo joined, then if D A be 

 equal to D B or D c, tho anglo at A U a right angle. 



LESSONS IN CII V. XXXL 



FERME.N 



The Sugan. Tho Faccharine group occupies a prominent 

 place in Organic Chemistry. It is closely allied to the amyla- 

 ceous group, for, as we have already seen, starch can be con- 

 verted into sugar ; moreover, the alcohols and their allies owe. 

 their existence to tho decomposition of the members of thix 

 group. The sugars all possess a charactcriatio sweet taste. 



