PROPOSITION C. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles must be equan in all respects. D CE Let the three sides of the As ABC, DEF be equal, each to each, that is, AB=DE, AC=DF, and BC=EF. Then must the triangles be equal in all respects. Imagine the 9 DEF to be turned over and applied to the A ABC, in such a way that EF coincides with BC, and the vertex D falls on the side of BC opposite to the side on which A falls ; and join AD. CASE I. When AD passes through BC. that is, Then in A ABD, : BD=BA, .. _BAD= BDA, I. A. And in A ACD, :: CD=CA, .. _ CAD= 2 CDA, I. A. .:: sum of 28 BAD, CAD=sum of 28 BDA, CDA, Ax. 2. BAC= = LBDC. Hence we see, referring to the original triangles, that < BAC=LEDF. .., by Prop. 4, the triangles are equal in all respects. = 19 CASE II. When the line joining the vertices does not pass through BC. B D Then in A ABD, :: BD=BA, .. 2 BAD= . BDA, I. A. 2 And in 1 ACD, :: CD=CA, .. _ CAD= 2 CDA, I. A. Hence since the whole angles BAD, BDA are equal. and parts of these CAD, CDA are equal. .:: the remainders BAC, BDC are equal. Ax. 3. Then, as in Case I., the equality of the original triangles may be proved. CASE III. When AC and CD are in the same straight line. Then in A ABD, :: BD-BA, BAD= BDA, I. A. that is, _BAC= BDC. Then, as in Case I., the equality of the original triangles may be proved. Q. E. D. PROPOSITION C. THEORF If two triangles have the three sides three sides of the other, each to each, th, in all respects. B Let the three sides to each, that is, AB Then must the t ve the given angle. In AC make AE=AD, and join DE. any pt. D. CASET I. 1. chi DE, on the side remote from A, describe an parilat. A DFE. Join AF. Then AF will bisect BAC. For in As AFD, AFE, I. c. · AD=AE, and AF is common, and FD=FE, ...DAF= LEAF, that is, BAC is bisected by AF. Q. E. F. Ex. 1. Shew that we can prove this Proposition by means of Prop. 1V. and Prop. A., without applying Prop. C. Ex. 2. If the equilateral triangle, employed in the construction, be described with its vertex towards the given angle ; shew that there is one case in which the construction will fail, and two in which it will hold good. NOTE.— The line dividing an angle into two equal parts is pulled the BISECTOR of the angle. Let AB be the gi I to AB. itting AB Bisect - ACB by the st. line CD meeting AB in D; then AB shall be bisected in D. I. 10. For in As ACD, BCD, .:. AD=BD; 1.4, Q. E. F. Ex. 1. The straight line, drawn to bisect the vertical angle of an isosceles triangle, also bisects the base. Ex. 2. The straight line, drawn from the vertex of an isosceles triangle to bisect the base, also bisects the vertical angle. Ex. Produce a given finite straight line to a point, such that the part produced may be one-third of the line, which is made up of the whole and the part produced. PROPOSITION IX. PROBLEM. To bisect a given angle. A B Let BAC be the given angle. It is required to bisect 2 BAC. In AB take any pt. D. In AC make AE=AD, and join DE. On DE, on the side remote from A, describe an equilat. A DFE. I. 1. Join AF. Then AF will bisect . BAC. For in AS AFD, AFE, :: AD=AE, and AF is common, and FD=FE, ... DAF= LEAF, that is, .BAC is bisected by AF. Q. E, F. Ex. 1. Shew that we can prove this Proposition by means of Prop. IV. and PROP. A., without applying Prop. C. Ex. 2. If the equilateral triangle, employed in the construction, be described with its vertex towards the given angle ; shew that there is one case in which the construction will fail, and two in which it will hold good. NOTE.— The line dividing an angle into two equal parts is called the BISECTOR of the angle. I. c. |