PROPOSITION XII. THEOREM 102. Two right triangles are equal if the hypotenuse and a leg of one are equal to the hypotenuse and a leg of the other. Given two right A ABC and DEF having the hypote nuse AB hypotenuse DE, and BC= EF. Proof. Place the ▲ ABC so that BC shall coincide with its equal, EF, and A fall on the opposite side of EF from D, at A'. Geom. Ax. 2. Then A'F and FD will form a straight line, A'F'D, Art. 74. (if two adj. 4 are together equal to two rt. &, their ext. sides form one and the same straight line). (two right ▲ are equal if the hypotenuse and an acute 4 of one are equal to the hypotenuse and an acute of the other). PROPOSITION XIII. THEOREM 103. An exterior angle of a triangle is greater than either opposite interior angle. FC. Proof. Let E be the middle point of the line BC. Draw AE and produce it to F, making FE-AE. Draw Then, in the A AEB and FEC, AE L BEA / FEC (being vertical 4). Art. 78. ΔΑΕΒ - Δ FEC, Art. 96. (two ▲ are equal if two sides and the included ▲ of one are equal, respectively, to two sides and the included 4 of the other). But .. LABEL FCE, (being homologous & of equal ▲). ZBCD is greater than FCE, (the whole is greater than any of its parts). Substituting ABE for its equal FCE, Ax. 7. Ax. 8. LBCD is greater than ZABE, that is, than ZABC. Similarly, by drawing a line from B through the midpoint of AC and by producing BC through C to a point H, it may be shown that ZACH (= LBCD) is greater than BAC. Q. E. D. Ex. On a given line, as base, construct exactly an isosceles triangle each of whose legs equals half a given line. PROPOSITION XIV. THEOREM 104. If two sides of a triangle are unequal, the angles opposite are unequal, and the greater angle is opposite the greater side B D AL Given the side BC > side AB in the ▲ ABC. To prove BAC greater than ≤ C. Proof. On the side BC take BD equal to AB and draw AD. Then, in the isosceles AABD, Lr= Ls, Art. 99. (in an isosceles ▲ the Ls opposite the equal sides are equal). (an ext. 4 of a ▲ is greater than either opposite int. Z). Much more, then, is BAC (which is greater than Ls) greater than ≤C, Ax. 12. (if, of three quantities, the first is greater than the second, and the second is greater than the third, then the first is greater than the third). Q. E. D. 105. NOTE. The essential steps of the above proof may be arranged in a single statement, thus: [BAC > < r = Ls > LC :. ≤ BAC is greater than C. Ex. 1. Which is the longest side of a right triangle? of an obtuse triangle? Ex. 2. Construct exactly an equilateral triangle, each of whose sides is half a given line. PROPOSITION XV. THEOREM (CONVERSE OF PROP. XIV) 106. If two angles of a triangle are unequal, the sides opposite are unequal, and the greater side is opposite the greater angle. B AL Given A greater than C in the AABC. To prove BC > AB. Proof. BC either equals AB, or is less than AB, or is greater than AB. C, Art. 99. for, if it did, ZA would equal But this is contrary to the hypothesis. Also BC cannot be less than AB, for, if it were, A would be less than C, Art. 104. (if two sides of a ▲ are unequal, the & opposite are unequal, and the greater is opposite the greater side). This is also contrary to the hypothesis. .. BC > AB, (for it neither equals AB, nor is less than AB). Q. E. D. Ex. 1. Draw a triangle the altitude of which falls on the base produced. What kind of a triangle is this? Ex. 2. Draw a triangle the altitude of which coincides with one side. What kind of a triangle is this? Ex. 3. By exact use of the ruler and compasses, draw a perpendicular to a given line from a given point without the line, PROPOSITION XVI. THEOREM 107. If two triangles have two sides of one equal, respectively, to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given the A ABC and DEF in which AB= DE, BC=EF, and ZABC is greater than ZE. Proof. Place the ▲ DEF so that the side DE coincides with its equal, the side AB, and F takes the position F. Geom. Ax. 2. Let the line BH bisect the FBC and meet the line AC at H. Draw FH. Then, in the AFBH and BHC, F"B=BC, Hyp. But :. F'H=CH, (homologous sides of equal A). (the sum of any two sides of a ▲ is greater than the third side). Substituting for HF its equal HC, AH + HC, or AC > AF". . AC > DF. Ax. 8. Ax. 8. Q. E. D. Ex. 1. Draw a figure for Prop. XVI in which the sides and angles are of such a size that I falls within the triangle ABC, Ex. 2. Draw another figure in which F falls on the side 40, |