PROP. 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 equal in all respects. 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 DEF to be turned over and applied to the ▲ 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. Then in AABD, ·: BD=BA, :. L BAD= L BDA. Prop. A. And in AACD, : CD=CA, :. LCAD L CDA. = Prop. A. ... sum of 4 8 BAD, CAD=sum of 4 8 BDA, CDA, Ax. 2. Hence we see, referring to the original triangles, that L BAC LEDF. .., by Prop. 4, the triangles are equal in all respects. CASE II. When the line joining the vertices does not pass through BC. B = Then in ▲ ABD, ·: BD=BA, :. ▲ BAD= L BDA. And in ▲ ACD, ·: CD=CA, :. L CAD LCDA. Hence since the whole angles BAD, BDA are equal, and parts of these CAD, CDA are equal, .. the remainders BAC, BDC are equal. Then, as in Case I., the equality of the original triangles may be proved. Ax. 3. CASE III. When AC and CD are in the same straight line. B A Then in ▲ ABD, :: BD=BA, :. L BAD= ▲ BDA, that is, BAC= L BDC, Then, as in Case I., the equality of the original triangles may be proved. Q. E. D. 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. ▲ DFE. Join AF. Then AF will bisect BAC. For in As AFD, AFE, :: AD=AE, and AF is common, and FD=FE, :. LDAF LEAF, that is, BAC is bisected by AF. I. C. 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. PROPOSITION X. PROBLEM. To bisect a given finite straight line. A Let AB be the given st. line. It is required to bisect AB. On AB describe an equilat. ▲ ACB. Bisect ACB by the st. line CD meeting AB in D; I. 9. then AB shall be bisected in D. For in As ACD, BCD, * AC= BC, and CD is common, and 4 ACD= 4 BCD, .. AD=BD; .. AB is bisected in D. I. 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. 3. 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 XI. PROBLEM. To draw a straight line at right angles to a giren straight line from a given point in the same. A Let AB be the given st. line, and Ca given pt. in it. Join FC. FC shall be to AB. For in As DCF, ECF, :: DC=CE, and CF is common, and FD= FE, Ex. 1. Shew that in the diagram of Prop. IX. AF and ED intersect each other at right angles, and that ED is bisected by AF. Ex. 2. If O be the point in which two lines, bisecting AB and AC, two sides of an equilateral triangle, at right angles, meet; shew that OA, OB, OC are all equal. Ex. 3. Shew that Prop. XI. is a particular case of Prop. IX. |