PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal, then the triangles shall be equal in all respects; that is to say, their bases or third sides shall be equal, and their remaining angles shall be equal, each to each, namely those to which the equal sides are opposite; and the triangles shall be equal in area. A A Let ABC, DEF be two triangles, in which the side AC is equal to the side DF, and the contained angle BAC is equal to the contained angle EDF. Then (i) the base BC shall be equal to the base EF; the angle ABC shall be equal to the angle DEF; (iii) the angle ACB shall be equal to the angle DFE; (iv) the triangle ABC shall be equal to the triangle DEF in area. Proof. If the triangle ABC be applied to the triangle DEF, Hyp. therefore the point B must coincide with the point E. And because AB falls along DE, and the angle BAC is equal to the angle EDF, Hyp. therefore AC must fall along DF. And because AC is equal to DF, Hyp. therefore the point C must coincide with the point F. Then since B coincides with E, and C with F, therefore the base BC must coincide with the base EF; for if not, two straight lines would enclose a space; which is impossible. Ax. 10. Thus the base BC coincides with the base EF, and is therefore equal to it. Ax. 8. And the remaining angles of the triangle ABC coincide with the remaining angles of the triangle DEF, and are therefore equal to them; namely, the angle ABC is equal to the angle DEF, and the angle ACB is equal to the angle DFE. And the triangle ABC coincides with the triangle DEF, and is therefore equal to it in area. That is, the triangles are equal in all respects. Ax. 8. Q.E.D. NOTE. The sides and angles of a triangle are known as its six parts. A triangle may also be considered in regard to its area. Two triangles are said to be equal in all respects, or identically equal, when the sides and angles of one are respectively equal to the sides and angles of the other. We have seen that such triangles may be made to coincide with one another by superposition, so that they are also equal in area. [See Note on Axiom 8.] [It will be shewn later that triangles can be equal in area without being equal in their several parts; that is to say, triangles can have the same area without having the same shape.] EXERCISES ON PROPOSITION 4. 1. ABCD is a square: prove that the diagonals AC, BD are equal to one another. 2. ABCD is a square, and L, M, and N are the middle points of AB, BC, and CD: prove that (i) LM =MN. (iii) AN=AM. (ii) AM DM. (iv) BN DM. [Draw a separate figure in each case. 3. ABC is an isosceles triangle: from the equal sides AB, AC two equal parts AX, AY are cut off, and BY and CX are joined. Prove that BY=CX. 4. ABCD is a quadrilateral having the opposite sides BC, AD equal, and also the angle BCD equal to the angle ADC: prove that BD is equal to AC. PROPOSITION 5. THEOREM. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles on the other side of the base shall also be equal to one another. B E Let ABC be an isosceles triangle, in which and let the straight lines AB, AC be produced to D and E. Construction. In BD take any point F; and from AE cut off a part AG equal to AF. Join FC, GB. Proof. Then in the triangles FAC, GAB, Because FA is equal to GA, and AC is equal to AB, I. 3. Constr. Нур. also the contained angle at A is common to the two triangles: therefore the triangle FAC is equal to the triangle GAB in all respects; that is, the base FC is equal to the base GB, I. 4. and AB, a part of AF, is equal to AC, a part of AG; Hyp. therefore the remainder BF is equal to the remainder CG. Because and FC is equal to GB, Proved. Proved. also the contained angle BFC is equal to the contained angle CGB, Proved. therefore the triangle BFC is equal to the triangle CGB in all respects; so that the angle FBC is equal to the angle GCB, I. 4. Now it has been shewn that the angle ABG is equal to the angle ACF, and that the angle CBG, a part of ABG, is equal to the angle BCF, a part of ACF; therefore the remaining angle ABC is equal to the remaining angle ACB; Ax. 3. and these are the angles at the base of the triangle ABC. Also it has been shewn that the angle FBC is equal to the angle GCB; and these are the angles on the other side of the base. Q.E.D. COROLLARY. Hence if a triangle is equilateral it is also equiangular. NOTE. The difficulty which beginners find with this proposition arises from the fact that the triangles to be compared overlap one another in the diagram. This difficulty may be diminished by detaching each pair of triangles from the rest of the figure, as shewn in the margin. F C B B F PROPOSITION 6. THEOREM. If two angles of a triangle be equal to one another, then the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. B Let ABC be a triangle, in which Construction. For if AC be not equal to AB, Proof. Then in the triangles DBC, ACB, Because DB is equal to AC, and BC is common to both, I. 3. Constr. also the contained angle DBC is equal to the contained angle ACB; Hyp. therefore the triangle DBC is equal to the triangle ACB I. 4. in area, Therefore AB is not unequal to AC; Q.E.D. COROLLARY. Hence if a triangle is equiangular it is also equilateral. |