12 



OF SPACE. 



99. Theoeem. Of two triangles on the 

 sartie base, the sides of the interior contain 

 the greater angle. 

 E 



Produce AB to C, then z.ABD>ACD 

 (07), and /. ACD> AEC, therefore much 

 more ABD>AED. 



'B 



A T) 



100. Problem. To make a triangle, hav- 

 ing its sides equal to three given right lines, 

 every one of them being less than the sum 

 of the other two. 



Take AB equal to one of the lines, 

 and on the centres A and B describe two 

 circles with radii equal to the other two 

 A B lines; draw AC and BC to the intersec- 



tion C, and ABC will be the triangle required. 



101. Problem. At a given point in a right 

 line, to make an angle equal to a given angle. 



» /" p In the lines forming the given angle 



ABC, take any two points, A and C, 

 join AC, and taking DEziBC, make 

 S C D E the triangle DEF, having DF=BA 

 and FE=AC (too), then /.FDE=:ABC (sg). 



102. Theorem. If two triangles have two 

 angles and a side respectively equal, the 

 whole triangles are equal. 



p j;^ Let the equal sides be AB and 



CD, intervening between the equal 

 angles, then if on AB a triangle 

 equal to CDE be supposed to be 

 A Cf 15 C I) constructed, the points A and B, 



and the angles at A and B being the same in this triangle 

 and in ABF, the sides must coincide both in position and 

 in length ; therefore ABF=:CDE. 



If the equal sides are AF and CE, opposite to equal 

 angles, then ABz:CD, and the whole triangles are equal. 

 For if AB is not equal to CD, let it be the greater, and let 

 .AGzzCD ; then, by what has been demonstrated, the 

 triangle AFG=:CED, and z.AGF=CDE=rABF, by the 

 supposition-; but AGF>ABF (9"), which is impossible. 



103. Theorem. The shortest of all right 

 lines that can be drawn from a given point to 

 a given right line is that which is perpendi- 

 cular to the line, and others arc shorter as 

 they are nearer to it. 



B. 



D 



T) 



E 



Let AB be perpendicular to CD, then 

 AB is shorter than AD . Produce AB, 

 take BE=AB, and join DE ; then the 

 triangle ABD=:EBD (86), and AD=: 

 DE. But AB+BE or 2AB is less than 

 AD+DE or 2AD (79), therefore AB 

 < AD (is). In a similar manner 2AD 

 <2AF(80), and AD<AF. 



104. Theorem. If a right line cutting^ 

 two others, makes the alternate angles equal, 

 the two lines are parallel. 



If .ilABC=ADE; BC 



and DE are parallel : for 



if they meet, as in F, they 

 will form a triangle BDF, ^ 

 and ^ADE>ABC (97). 



105. Theorem. A right linCKutling two 

 parallel lines, makes equal angles with them. , 



Let AB cut the parallels BC, ^ 



DE; then if /.ABC is not equal Bx^^ C 



to ADE, let it be equal to ADF, ^y^ jf 



then BC and DFare parallel (104), 

 andDE, which cuts DF, will also, 

 if produced, cut BC (83), contra- 

 rily to the supposition. 



106. Theorem. Right lines, parallel to the 

 same line, are parallel to each other. 



Let AB and CD be parallel to' 

 EF ; draw GHI cutting them all, -2^ 

 then z.K.GB=KIF (los), and 

 aKHD=:KIF, therefore A KGB 

 =K.HD, and AB|1CD (104). 



107. Problem. Through a given point to 

 draw a right line parallel to a given right hne. 



From A draw, at pleasure, AB, meet- A D 



ingBCin B, and make /.BAD=ABC 



(I0l),then ADllCB (104). ,.. 



C/ li 



108. Theorem. The angles of any trian"-le 

 taken together, are equal to two right angles.'X^ 



Produce AB'to C, and draw BD paral- 

 leltoAE. Then /.EBDrzAEB (IC5), ^ 



and /.DBC':=EAB; therefore the exter- 

 nal angle EBC is equal to the sum of the 

 internal opposite angles, AEB, EAB, and 



B 



