PROPOSITION XXVI. THEOREM. If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely either the sides adjacent to the equal angles in each, or the sides opposite to them; then shall the other sides be equal, each to each, and also the third angle of the one equal to the third angle of the other, Let ABC, DEF be two As which have 2 s ABC,BCA = Ls DEF, EFD, each to each, namely ABC to DEF and BCA to EFD; also one side equal to one side. First let those sides be equal which are adjacent to the angles that are equal, namely BC to EF. Then the other sides shall be equal, each to each, namely, AB to DE and AC to DF, and the third L BAC to the third EDF. First case. Constr. For if AB be not = DE, one of thein must be greater than the other ; let AB be the greater. Make BG=ED, (i. 3) and join GC. Dem. Then in the As GBC, DEF, :: GB= DE, and BC= EF, (hyp.) and LGBC= L DEF; (hyp.) .. base GC = base DF, and a GBC = A DEF; (i. 4) :: L GCB = L DFE; but L ACB = LDFE; (hyp.) .. also _ GCB = L ACB; the less equal to the greater, which is impossible; AB is not unequal to DE, that is, it is equal to it. Hence in As ABC, DEF, :: AB=DE, and BC = EF, and L ABC= _ DEF:.. base AC = base DF, and third L BAC = third _ EDF. (i. 4) Secondly, let sides which are opp. one of the = Ls in each A be equal to one another, namely, AB = DE. In this case likewise the other sides shall be equal, AC to DF, and BC to EF, and the third L BAC to third LEDF. Second case. Constr. For if BC be not = EF, one of them must be greater than the other. Let BC be greater than EF; make BH = EF, (i. 3) and join AH. Dem. Then in As ABH, DEF, :: AB = DE, BH = EF, and L ABH= L DEF; (hyp.) .. base AH = base DF and A ABH = A DEF, (i. 4) ..L BHA = _ EFD; but L EFD= L BCA (hyp.);:: LBHA = L BCA, (ax. 1) that is, the ext. L of A AHC= int. opp. BCA; which is impossible ; (i. 16) .: BC is not unequal to EF, that is BC= EF. Hence in As ABC, DEF; :: AB=DE, BC = EF, and L ABC= L DEF; .. base AC= base DF, (i. 4) and third L BAC = third _ EDG. Therefore, if two triangles, &c. Q. E. D. (This is the third case of the equality of two As, the two others being props. 4 and 8. Observe that the two cases (1) when the equal sides are adjacent, (2) when they are opposite the equal angles, are quite independent. It is a very common error to prove the equality of two other sides, each to each, and then to stop with the idea that the demonstration is completed. This prop. ends the first section of the first Book.] 1. From a given point draw a line making equal angles with two given lines. 2. In a given straight line to find a point such that the perpendiculars from it upon two given straight lines are equal. 3. If two right-angled triangles have two sides containing an acute angle of the one equal to two sides containing an acute angle of the other, each to each; show that the triangles are equal in all respects. 4. If the straight line which bisects the vertical angle of a triangle also bisects the base, the triangle is isosceles. 5. In the figure of prop. 5, if FC, BG intersect in H, show that FH, GH are equal. 6. In the same figure, show that AH bisects the angle BAC 7. The perpendicular is the shortest line that can be drawn from a given point to a given line, and that which is nearer to the perpendicular is less than one more remote; and from the same given point there can be drawn only two equal lines to the given line, one on each side of the shortest line. PROPOSITION XXVII. THEOREM. If a straight line, falling on two other straight lines, make the alternate angles equal to one another, these two straight lines shall be parallel. Let the st. line EF, which falls on the two st. lines AB, CD, make the alternate LS AEF, EFD equal to one another. Then shall AB be || to CD. For, if not, AB and CD being produced will meet either towards A and C, or towards B and D. Let them be produced and meet towards B and D, at the pt. G. Dem. Then GEF is A, and its ext. LAEF is greater than the int. opp. _EFG; (i. 16) But LAEF = EFG; (hyp.) :: LAEF is greater than, and equal to, _ EFG, which is impossible. :. AB, CD being produced do not meet towards B, D. In like manner, it may be demonstrated that they do not meet when produced towards A, C. But those st. lines which being produced ever so far both ways do not meet, are 11 . (Def. 35) .. AB || CD. Therefore, straight lines which, &c. Q. E. D. [Alternate angles are the two angles which two straight lines make with another at its extremities, but on opposite sides of it.] 1. The quadrilateral whose diagonals bisect each other is a parallelogram. 2. A line joining two parallel straight lines is bisected: show that any other straight line drawn through the point of bisection to meet the parallel lines will be bisected in that point. 3. A line bisecting the vertical angle of a triangle meets the base in D; DE, DF parallel to the sides meet them in E, F. Prove that DE, DF are equal. 4. A line is drawn through the vertex of an isosceles triangle so as to make equal angles with the sides meeting in the vertex ; prove that this line must be either parallel to the base or perpendicular to it. |