= angle EGF BAC, therefore angle EDF = BAC. Consequently triangle ABC triangle DEF, (126.) 2. Right angles, (fig. 89. 2.) By supposition ED EG, therefore angle DG. Angle AG, therefore A D; consequently triangle ABC = triangle DEF, (126.) = = = 3. Obtuse angles, (fig. 89. 3.) By supposition ED= EG and FD=FG, therefore angle EDG=EGD, and angle FDG FGD. Consequently angle EDGFDG= angle EDF = EGD FGD= EGF. Now angle BAC EGF, therefore angle BAC = = the triangle ABC=DEF, (126.) EDF; and In each case equal angles are opposite to equal sides. 139. In every triangle the greater side is opposite to the greater angle, (fig. 90.) If AC > AB, then is angle Bangle C. Dem. In the side AC take AD=AB and draw BD. Angle ABD=ADB, (129.) But angle ADB > ACB, (124,) much more is angle ABC=ABD+DBC> ACB. 140. Cor. In every triangle whose sides are not equal, the greatest side is opposite to the greatest angle, and the least side to the least angle. 141. In every triangle the greatest angle is opposite to the greatest side, (fig. 90.) If angle B>C, then is ACAB. Dem. Draw BD, making the angle CBD = angle Because angle CBD=BCD, therefore BDCD, (133;) DA=DA; and therefore BD+DA=CD+ DA. But BD+ DA> BA, consequently CD+DA— CABA. 142. In every triangle whose angles are not equal, the least angle is opposite to the least side, the mean angle to the mean side, and the greatest angle to the greatest side. 143. In the two triangles ABC and DEF, (fig. 109,) let the sides AB and AC be respectively equal to the sides DE and DF, but the angle EDF be greater than the angle BAC; then will the side EF be greater than the side BC. Dem. Let the triangle ABC be placed upon the triangle DEF, so that AB shall coincide with DE; but as angle BAC is less than EDF, the side AC will not fall upon DF, but will fall within it, suppose in the position DG; BC will lie in the position EG, and triangle DEG will be equal to triangle ABC. Draw FG. In the isosceles triangle DFG, angle DGF = angle DFG, (129.) It is self-evident that DFG is greater than EFG; therefore DGF is greater than EFG; much more is EGF greater than EFG; therefore the side EF is greater than the side EG, or its equal BC, (138.) 144. The converse of this proposition is true, viz. : that if the sides AB and AC are respectively equal to the sides DE and DF, but the side EF is greater than the side BC, then will the angle EDF be greater than the angle BAC. For angle EDF cannot be equal to angle BAC, for then the two triangles will be equal, and the side EF-BC, which is contrary to the supposition; neither can angle EDF be less than angle BAC, for then the side EF would be less than the side BC, which is also contrary to the supposition; angle EDF must therefore be greater than angle BAC. 145. In every quadrilateral the sum of all the angles is equal to 4 R. A. Dem. By a diagonal the quadrilateral is divided into 2 triangles. The sum of the angles of each triangle is equal to 2 R. A.; consequently the sum of the angles of both triangles, that is, of the quadrilateral, is equal to 4 R. A. 146. Every parallelogram ABCD is divided by a diagonal DB into two equal triangles, (fig. 91.) Dem. AB is parallel to DC, therefore angle ABD =BDC, (110;) AD is parallel to BC, therefore angle BDA=DBC; BD=BD; therefore triangle ABD= CBD, (127.) 147. Cor. In equal triangles equal sides are opposite to equal angles; consequently AD=CB, and AB =CD; that is, in every parallelogram the sides opposite to each other are equal. Each part of the angle B is equal to a part of the angle D, therefore the entire angle B is equal to the entire angle D; the angle C=A; that is, in every parallelogram the opposite angles are equal. 148. Equivalent figures are such as have equal surfaces. Two figures may be equivalent, although dissimilar in form; thus a circle may be equivalent to a square. 149. The altitude of a parallelogram is a line drawn from one side perpendicular to the opposite side considered as the base, or to that side produced if necessary. The altitude of a triangle is a perpendicular drawn from the vertex of an angle to the opposite side taken as a base; as CD is the altitude of triangle ABC, (fig. 86.) 150. Two parallelograms having the same or equal bases and an equal altitude, are equivalent. Dem. Upon one of the sides of the parallelogram ABCD (fig. 92,) construct a second parallelogram ABEF, of which the side opposite to AB shall be in DC produced, that is, which shall have an equal altitude with ABCD. We are to prove that ABEF is equivalent to ABCD. AB=DC; AB=FE, (145,) therefore DC FE; CF-CF; therefore DC + CF =DF FE+CFEC. Again, AD BC, and AF -BE, therefore triangle ADF BCE, (138.) = = = Triangle OCF OCF, therefore triangle ADFOCF = ADCO is equal to triangle BCE — OCF = BOFE. Triangle AOBAOB, therefore ADCO+ AOB ABCD BOFE+ AOB = ABEF. = Let the parallelograms have not the same but equal bases, and have equal altitudes; we can suppose one to be placed upon the other, so that the bases shall coincide, and then the case becomes identical with the preceding. 151. Cor. Every oblique parallelogram is equivalent to a right parallelogram of an equal base and an equal altitude. 152. Every triangle is equivalent to one half of a parallelogram of equal base and equal altitude. Dem. Let BC= EF, and AG be parallel to BF. Draw CH parallel to BA. Parallelograms ABCH and DEFG are equivalent, (150.) Triangle ABC=} 153. Cor. Two triangles of equal bases and equal altitudes are equivalent, because each is the half of a parallelogram having an equal base and an equal altitude. 5. POLYGONS IN GENERAL. 154. The sum of all the angles of a rectilineal figure is equal to twice as many right angles, wanting 4, as the figure has sides. 2 Dem. From any point within the figure draw straight lines to the vertex of each angle. Thus the figure will be divided into as many triangles as it has sides. The sum of all the angles of each triangle R. A., therefore the sum of all the angles of all the triangles is equal to 2 R. A. multiplied by the number of sides in the figure. But the angles (equal to 4 R. A.) about the point within the figure do not belong to the figure; therefore we must deduct 4 R. A. For example, the angles of a triangle together = (3 × 2) — 4 = 2 R. A. (4X2)-4= 4 R. A. (5 × 2)-4= 6 R. A. (20 × 2)-4=36 R. A. 155. In a regular polygon all the angles are equal, therefore each angle of a regular triangle |