Let ABC be a triangle, having the angle ABC equal to the angle ACB; A B Construction. If AC is not equal to AB, one of them must be greater than the other; and from BA the greater cut off BD equal to CA the less; (prop. 3) join CD. (post. 1) Demonstration. Then in the triangles DBC, ACB; because DB is equal to AC, and BC is common to both triangles, and the contained angle DBC is equal to the contained angle ACB, therefore the triangle DBC is equal to the triangle ACB: (prop. 4) that is, the less triangle is equal to the greater, which is absurd. Therefore AC is not unequal to AB, that is, AC is equal to AB. Q.E.D. COROLLARY.-Hence every equiangular triangle is also equilateral. Exercises. 1. The angles at the base of an isosceles triangle ABC are bisected by two straight lines BD, CD; shew that DCB is an isosceles triangle. 2. If in the figure on page 14, BG and CF intersect at H, shew (1) that HF and HG are equal to each other; (2) that AH bisects the angle BAC. 3. If G is a point within an equilateral triangle such that the straight lines GA, GB, GC bisect the three angles of the triangle; shew that the point & is equidistant from the angular points of the triangle. 4. ABC is a triangle in which the angle B is double of the angle C; shew that the line which bisects the angle B cuts AC in D, so that BD is equal to CD. 5. Two angles of an equilateral triangle ABC are bisected by straight lines AG, BG: shew that CG bisects the third angle. C [NOTE. This proposition is only used by Euclid as leading to the next, which may however be proved independently of the seventh; the student is therefore advised to omit this proposition on his first reading of Euclid.] Upon the same base, and on the same side of it, there cannot be two triangles having their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity. If it be possible, upon the same base AB, and on the same side of it, let there be two triangles ACB, ADB having their sides CA, DA, which are terminated in one extremity A of the base, equal to one another, and likewise their sides CB, DB, which are terminated in the other extremity B. First.--Let the vertex of each of the triangles fall without the Because in the triangle ACD, the side AC is equal to the side AD, therefore the angle ACD is equal to the angle ADC; (prop. 5) but the angle ACD is greater than the angle BCD, (ax. 9) therefore the angle ADC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD. But since in the triangle BDC, the side BD is equal to the side BC, therefore the angle BDC is equal to the angle BCD. (prop. 5) Hence the angle BDC is both greater than and equal to the angle BCD; which is impossible. Secondly.-Let the vertex D of one triangle ADB be within the other triangle ACB. A E B Construction. Join CD. Produce AC to E, and AD to F. Demonstration. Because in the triangle ACD, the equal sides AC, AD are produced to E and F, therefore the angle ECD is equal to the angle FDC; (prop. 5) but the angle ECD is greater than the angle BCD, (ax. 9) therefore the angle FDC is also greater than the angle BCD; much more then is the angle BDC greater than the angle BCD. But since in the triangle BCD, the side BD is equal to the side BC, therefore the angle BDC is equal to the angle BCD. (prop. 5) Hence the angle BDC is both greater than and equal to the angle BCD; which is impossible. Thirdly. Let the vertex D of one triangle ADB fall on а side of the triangle ACB. In this case it is evident that AD is not equal to AC. (ax. 9) Therefore in no case can there be two triangles ACB, ADB, on the same side of AB, having the sides AC, AD equal to each other, and likewise the sides BC, BD. Q.E.D. Exercises. 1. Shew that there can be two triangles ACB, ADB placed as above, in which BC is equal to BD. 2. Shew that if ADB be on the other side of AB, then AC, AD may be equal to each other, and likewise BC, BD. 3. In the figure of the first case, if AD and BC are equal, and make equal angles with AB, shew that AC and BD are also equal and make equal angles with AB. PROPOSITION VIII. THEOREM. If two triangles have the three sides of the one equal to the three sides of the other, each to each; the two triangles shall be equal in all respects. Let ABC, DEF be two triangles, in which AB is equal to DE, AC to DF, and BC to EF; then shall the triangles ABC, DEF be equal in all respects. Construction. Conceive the triangle ABC to be placed so that the base BC coincides with the base EF, but having the vertex A on the opposite side of EF from D; let A'EF represent this new position of ABC. Join DA'. Demonstration. Because in the triangle EDA', the side ED is equal to the side EA', therefore the angle EDA' is equal to the angle EA'D: (prop. 5) again, in the triangle FDA', since the side FD is equal to the side FA', therefore the angle FDA' is equal to the angle FA'D: (prop. 5) therefore the whole angle EDF is equal to the whole angle EA'F; (ax. 2) hence the triangles DEF, A'EF must be equal in all respects. (prop. 4) But the triangle A'EF is the triangle ABC in another position, therefore the triangles ABC, DEF are equal in all respects. Secondly.-Let DA' intersect EF at one of its extremities. Because the triangle EDA' is isosceles, therefore the angle EDA' is equal to the angle EA'D; (prop. 5) hence the triangles DEF, A'EF are equal in all respects. (prop. 4) But the triangle A'EF is the triangle ABC in another position, therefore the triangles ABC, DEF are equal in all respects. Thirdly. Let DA' intersect EF produced in some point G. Because the triangle EDA' is isosceles, therefore the angle EDA' is equal to the angle EA'D: (prop. 5) again because the triangle FDA' is isosceles, (prop. 5) therefore the angle FDA' is equal to the angle FA'D: therefore the angle EDF is equal to the angle EA'F; hence the triangles EDF, EA'F must be equal in all respects. (ax. 3) (prop. 4) But the triangle A'EF is the triangle ABC in another position, therefore the triangles ABC, DEF are equal in all respects. PROPOSITION VIII. THEOREM. Q.E.D. [NOTE. The following is Euclid's statement and proof of this proposition.] If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; then the angle which is contained by the two sides of the one shall be equal to the angle which is contained by the two sides equal to them, of the other. |