PROP. 23.-Problem.—At a given point in a given straight line to make an angle equal to a given rectilineal angle. Let A be the given point in the given straight line AB, and CDE the given rectilineal angle; it is required to make at A an angle equal to the angle CDE. ДД B Construction. In DC and DE take any points C and E. On AB describe the triangle AFG having its sides equal to the sides of the triangle CDE, namely, AF equal to DC, AG equal to DE, and FG equal to CE. The angle FAG shall be equal to the angle CDE. Prop. 22 Proof.-Because in the two triangles FAG, CDE Constr. Because and AG is equal to DE Constr. Constr. and the base FG is equal to the base CE, therefore by Prop. 8 the angle FAG is equal to the angle CDE. Therefore at the given point A in the given straight line AB the angle FAG has been made equal to the given rectilineal angle 1. At the point A in the straight line AB make another angle equal to the given angle CDE. 2. What previous Proposition is a special case of Proposition 22? 3. If the angle BAC of the triangle ABC be equal to the sum of the two angles ABC and ACB, show that the triangle ABC can be divided into two isosceles triangles. PROP. 24.-Theorem. -If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of the one greater than the angle contained by the two sides of the other; then, the base of that which has the greater angle shall be greater than the base of the other. Let ABC, DEF be two triangles, having AB equal to DE and AC to DF, but the angle BAC greater than the angle EDF; the base BC shall be greater than the base EF. A D G B C Construction. Of the two sides DE, DF, let DE be that which is not greater than DF, at the point D in the straight line DE and on the same side as DF make the angle EDG equal to the angle BAC, Prop. 23 and make DG equal to DF or AC. Join EG, GF. Proof.-Because in the two triangles ABC, DEG (AB is equal to DE, Because and AC is equal to DG, Constr. and the contained angle BAC is equal to the contained angle EDG, Constr. therefore by Prop. 4 the base BC is equal to the base EG. Again, because DF is equal to DG, therefore the angle DGF is equal to the angle DFG. Prop. 5 But the angle DGF is greater than its part the angle EGF, therefore the angle DFG is also greater than the angle EGF. Much more then is the angle EFG greater than the angle EGF. Because in the triangle EFG the angle EFG is greater than the angle EGF, therefore EG is greater than EF. Prop. 19 Q. E. D. PROP. 25.-Theorem.-If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; then the angle contained by the two sides of that which has the greater base shall be greater than the angle contained by the two sides of the other. Let ABC, DEF be two triangles, having AB equal to DE and AC to DF, but the base BC greater than the base EF; the angle BAC shall be greater than the angle EDF. Proof. Because, if the angle BAC be not greater than the angle EDF, BAC must be either equal to or less than EDF. But the angle BAC is not equal to the angle EDF; for then the base BC would be equal to the base EF, and it is not. Prop. 4 And the angle BAC is not less than the angle EDF; for then the base BC would be less than the base EF, and it is not. Prop. 24 Therefore the angle BAC is neither equal to nor less than the angle EDF, therefore the angle BAC must be greater than the angle EDF. Q. E. D. EXERCISE. ABCD is a quadrilateral figure having the side AD equal to the side BC, but the side CD greater than the side AB. Prove that the angle CAD is greater than the angle ACB, and the angle CBD greater than the angle ADB. PROP. 26.- Theorem.—If two triangles have two angles of the one equal to two angles of the other, each to each, and a side of the one equal to a side of the other, these sides being either adjacent or opposite to equal angles; then, the triangles shall be equal in all respects: that is, the remaining sides shall be equal, each to each; the third angle of the one shall be equal to the third angle of the other; and the triangles shall be equal in area. Let ABC, DEF be two triangles, having the angle ABC equal to the angle DEF, and the angle ACB equal to the angle DFE. CASE I. Let the side BC be equal to the side EF, these being sides adjacent to the equal angles; the triangle ABC shall be equal to the triangle DEF in all respects: that is, AB shall be equal to DE, AC shall be equal to DF; the third angle BAC shall be equal to the third angle EDF; and the triangle ABC shall be equal to the triangle DEF in area. A D AA Proof.-Because, if AB is not equal to DE, one of them must be the greater. If possible, let AB be the greater, and from AB cut off BG equal to DE. Join GC. Because in the two triangles GBC, DEF (GB is equal to DE, Because and BC is equal to EF, Prop. 3 Constr. Нур. and the contained angle GBC is equal to the contained therefore by Prop. 4 the triangles GBC and DEF are equal in all respects, therefore the angle GCB is equal to the angle DFE. But the angle ACB is equal to the angle DFE, therefore the angle GCB is equal to the angle ACB: that is, the part is equal to the whole, which is absurd. AA B E Hyp. Ax. 9 Therefore AB is not unequal to DE, that is, AB is equal to DE. Again, because in the two triangles ABC, DEF AB is equal to DE, Because and BC is equal to EF, Proved Hyp. and the contained angle ABC is equal to the contained therefore by Prop. 4 the triangle ABC is equal to the triangle DEF in all respects : that is, AC is equal to DF; the angle BAC is equal to the angle EDF; and the triangle ABC is equal to the triangle DEF in area. Q. E. D. CASE II. Let the side AB be equal to the side DE, these being sides opposite to equal angles; the triangle ABC shall be equal to the triangle DEF in all respects: that is, BC shall be equal to EF, AC shall be equal to DF; the third angle BAC shall be equal to the third angle EDF ; and the triangle ABC shall be equal to the triangle DEF in area. |