A E FIG. 211. DEM.-Let ACB and DEF be two triangles having AC = DF, CB = FE, and C> F; then is AB > DE. For, placing the side DF in its equal AC, since angle F< angle C, FE will fall within the angle ACB, as in CE. Then let the triangle ACE = the triangle DFE. BiEsect ECB with CH, and draw HE. The triangles HCB and HCE have two sides and the included angle of the one, respectively equal to the corresponding parts of the other, whence HE HB. Now AH + HE > AE; but AH + HE AH + HB = Therefore, AB > AE. Q. E. D. AB. = 296. COR. Conversely, If two sides of one triangle are respectively equal to two sides of another, and the third sides unequal, the angle opposite this third side is the greater in the triangle which has the greater third side. DEM.-If AC = DF, CB = FE, and AB > DE, angle C > angle F. For, if CF, the triangles would be equal, and AB = DE (284); and, if C were less than F, AB would be less than DE, by the proposition. But both these conclusions are contrary to the hypothesis. Hence, as C cannot be equal to F, nor less than F, it must be greater. PROPOSITION XI. 297. Theorem.-Two right angled triangles which have the hypotenuse and one side of the one equal to the hypotenuse and one side of the other, each to each, are equal. DEM.-In the two triangles ABC and DEF, right angled at B and E, let AC = DF, and BC = EF; then are the triangles equal. the perpendicular (141), DE = BA; and revolving CAB upon FB, the two triangles will coincide when CAB falls in the plane on the side D. Therefore, the triangles are equal. Q. E. D. PROPOSITION XII. 298. Theorem.-Two right angled triangles having the hypotenuse and one acute angle of the one equal to the hypotenuse and an acute angle of the other, are equal. DEM.-One acute angle in each being equal, the other acute angles are equal, since they are complements of the same angle (222). The case is, then, that of two angles (the acute angles in each), and their included side (the hypotenuse), and falls under (286). PROPOSITION XIII. 299. Theorem.-Two right angled triangles having a side and a corresponding acute angle in each equal, are equal. This also falls under (286). Let the student show why. OF QUADRILATERALS. PROPOSITION XIV. 300. Theorem.-Two quadrilaterals are equal when the following parts are equal, each to each, in both quadrilaterals, and similarly arranged: 1. The triangles into which either diagonal divides the quadrilaterals. 2. The four sides and either diagonal. 3. The four sides and one angle. 4. Three sides and the two included angles. 5. Three angles and any two sides, if the other two sides are nonparallel. DEM.-1. This case is demonstrated by applying one quadrilateral to the other. 2. This case is reduced to the former by (292). 3. Drawing the diagonal opposite the known angle, this case is reduced to the first by (284), and (292). 4. This is demonstrated by applying one quadrilateral to the other; or, draw. ing a diagonal, it may be reduced to case first by (284). 5. In this case the quadrilaterals are mutually equiangular by (255). If, then, the two sides are adjacent, by drawing the diagonal joining their extremi ties the case is brought under the first by (284), and (286). If, however, the FIG. 213. two known sides are non-adjacent, by producing the unknown sides two triangles are formed in each figure, which are mutually equal by (286), and the case comes under the axiom concerning equals subtracted from equals. [Let the pupil draw the figures and give the demonstrations in full form. Trapeziums should be used; although it should be seen in each case-except the fifth-that the truth applies to any other form of quadrilateral.] PROPOSITION XV. 301. Theorem.-Two parallelograms having two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, are equal. DEM.-Let AC and EC be two parallelograms, with AD = EH, AB = EF, and A = E; then are they equal. A B F FIG. 214. For, applying the angle E to A, since EH === AD, H will fall at D; and since EF = AB, F will fall at B. Now, through D but one line can be drawn parallel to AB; hence HG will fall in DC, and G will be found in DC, or in DC produced. In like manner, since but one parallel to AD can be drawn through B, FC must fall in BC, and G be found in BC, or in BC produced. Therefore, as G falls at the same time in DC and BC, it falls at C, and the parallelograms coincide. 302. COR.-Two rectangles of the same base and altitude are equal. OF POLYGONS. PROPOSITION XVI. 303. Theorem.-Two polygons of the same number of sides, having all the parts of the one except three angles, known to be respectively equal to the corresponding parts of the other, are equal. DEM.-If the two polygons AE* and A'E', have all the parts of the one equal to the corresponding parts of the other, each to each, except three angles; then are the polygons equal. * It is often more convenient to read a polygon by two letters, instead of all those at the vertices. ters coincide till we reach E' and E. Then will C'E' = CE, and the triangles GFE and G'F'E', having their corresponding sides respectively equal, are themselves equal, and the polygons coincide throughout. 2d. When two of the unknown angles are consecutive, and the third not consecutive with these, as G, E, D, and C', E', D'. From the angle which is not consecutive with the other two, draw diagonals to the other angles, as GE, GD, and C'E', C'D'. Now, G'A'B'C'D' can be applied to GABCD, and C'F'E' to GFE, in the ordinary way. Hence, the triangles C'E'D' and GED are mutually equilateral, and consequently equal. Therefore the polygons are equal. 3d. When no two of the three unknown angles are consecutive, as G, B, D, and C', B', D'. Join the unknown angles by diagonals, as GB, CD, BD, and C'B', C'D', B'D'. Now, polygon G'F'E'D' can be applied to GFED, D'C'B' to DCB, and C'A'B' to GAB in the ordinary way. Hence, the triangles C'D'B' and GDB are mutually equilateral, and consequently equal. Therefore the polygons are equal.* 304. COR.-Two quadrilaterals having their corresponding sides equal, and an angle in one equal to the corresponding angle in the other, are equal. 305. Theorem.-Two polygons of the same number of sides, having all the parts of the one except two angles and the included side, known to be respectively equal to the corresponding parts of the other, are equal. DEM.-Let the unknown parts be C, c, D, and C', c', D'. From any other two of the mutually equal angles, as G and G', draw the diagonals GC, GD, G'C', G'D', * Notice that in each case the unknown angles are to form the vertices of triangles, which the argument shows to be equilateral, and therefore equal. In Case 1st, we have to draw only one line in order to give the triangles, as two sides are sides of the polygon; in Case 2d, we have to draw two sides, and in Case 3d, three sides, for analogous reasons. A equals from G' to the unknown angles. Then the polygon G'F'E'D' can be applied in the ordinary way to GFED, f' being placed in f, etc. So also G'A'B'C' can be applied to GABC, begin'D' ning with g' in its equal g. Hence, angle F'G'D' = FGD, A'G'C' AGC; and, adding, F'G'D' + A'G'C' = FGD + AG C. Subtracting these G, we have C'G'D' CGD. Whence the triangles C'G'D' and CGD have two sides and their included angle equal in each, and are equal; therefore the polygons are equal in all their parts. A Α ́ 306. SCH.-When the unknown angles are both separated from the unknown side, the polygons may or may not be equal-the case is ambiguous. Thus, if C and E are the unknown angles and AH the unknown side, the polygons ABCDEFG, and A'B'C'D'EFG fulfill the conditions, but are not equal. By drawing CE, CA, and EH, the case is reHduced to that of two quadrilaterals having all the parts equal, each to each, except two angles and their non-adjacent side; in which case the quadrilaterals are not necessarily equal. So, also, when one of the unknown angles is adjacent to the unknown side and the other separated, the polygons may or may not be equal. Thus, let the unknown parts be D, c, G, and D', c', C'. From the separated angle draw the diagonals to the extremities of the unknown side, as GC, GD (or GD1), and G'C', C'D'. In the usual way G'A'B'C' can be applied to CABC, and C'F'E'D' to |