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PROFESSOR KELLAND ON THE LIMITS OF OUR KNOWLEDGE 



Prop. XIX. If two triangles ham the three angles of the one equal, each to each, to 

 the three angles of the other, the triangles are equal in every respect. 



Let ABC, DEF, be two triangles, which have the angle A = D, B=E, and C = r, 

 the triangles are equal in every respect. For their areas are equal (Prop. XVII.) 

 If, therefore, AB be greater than DE, AC must be less than DF. From AB cut 

 off AG=DE, and produce AC to H, making AH = DF ; the triangle AHG is equal 

 in every respect to DEF ; therefore, --::AHG =DFE = ACB, which is impossible. 

 The triangles ABC, DEF, are therefore equal in every respect. 



Props. XVL to XIX., again recall the properties of the 

 circle. They are in fact properties of spherical triangles 

 on the same sphere if the phrase angular defect be con- 

 verted into spherical excess. We may now apply the 

 (popular) idea of a circle to the triangle as in the accompanying figure. 



Prop. XX. Upon one of the sides of a given triangle to construct an isosceles triangle, 

 that shall have its area and the sum of its angles equal to those of the given 

 triangle. (This is Mr Meikle's proposition. It is introduced for the sake 

 of some corollaries). 



Let ABC be the given triangle. Bisect AB, AC, in D and E ; through DE 

 draw GDEH, and from A, B, C, draw perpen- 

 diculars AF, BG, CH, to GDEH. Bisect GH in 

 I, draw IK perpendicular to GH, and equal to AF, 

 join KB, KC; KBC is the triangle required. The 

 triangles ADF, BDG, are equal, as also the tri- 

 angles AEF, CEH (Euc. i. 26). Hence the figure 

 BGHC is equal to ABC in area, and the sum of the angles of ABC is equal to the 

 angles GBC, BCH. NowKI =AF = BG=CH. Therefore the triangles KIL and 

 BGL are equal, as also the triangles KIM, CHM, so that KM=MC; hence also 

 the triangle BKC is equal in area to the figure BGHC, and the sum of its angles 

 is the sum of the angles GBC, BCH ; therefore the triangles ABC, KBC, are 

 equal both in area and in the sum of their angles. 



Also because GL = LI, IM = MH, each of them is iGH ; therefore they are 

 equal, and BK=KC ; or BKC is an isosceles triangle. 



Cor. 1. The line which joins the points of bisection of the sides of a triangle 

 is parallel to the base. 



For BG = CH ; hence GH is parallel to BC (Prop. VI.) 



Cor. 2. If two equal triangles be upon the same base and on the same side of 



