Dem.—Because D is the centre of the circle EFG, DF is equal to DE (Def. xxxii.). And because DAB is an equilateral triangle, DA is equal to DB (Def. xxi.). Hence we have DF = DE, and DA = DB; and taking the latter from the former, the remainder AF is equal to the remainder BE (Axiom iii.). Again, because B is the centre of the circle ECH, BC is equal to BE; and we have proved that AF is equal to BE; and things which are equal to the same thing are equal to one another (Axiom i.). Hence AF is equal to BC. Therefore from the given point A the line AF has been drawn equal to BC. It is usual with commentators on Euclid to say that he allows the use of the rule and compass. Were such the case this Proposition would have been unnecessary. The fact is, Euclid’s object was to teach Theoretical and not Practical Geometry, and the only things he postulates are the drawing of right lines and the describing of circles. If he allowed the mechanical use of the rule and compass he could give methods of solving many problems that go beyond the limits of the “geometry of the point, line, and circle.”—See Notes D, F at the end of this work. Exercises. 1. Solve the problem when the point A is in the line BC itself. 2. Inflect from a given point A to a given line BC a line equal to a given line. State the number of solutions. PROP. III.—Problem. From the greater (AB) of two given right lines to cut off a part equal to (C) the less. Sol.—From A, one of the extremities of AB, draw the right line AD equal to C [ii.]; and with A as centre, and AD as radius, describe the circle EDF (Post. iii.) cutting AB in E. AE shall be equal to C. Dem.—Because A is the centre of the circle EDF, AE is equal to AD (Def. xxxii.), and C is equal to AD (const.); and things which are equal to the same are equal to one another (Axiom i.); therefore AE is equal to C. Wherefore from AB, the greater of the two given lines, a part, AE, has been out off equal to C, the less. Questions for Examination. 1. What previous problem is employed in the solution of this? 2. What postulate? 3. What axiom in the demonstration? 4. Show how to produce the less of two given lines until the whole produced line becomes equal to the greater. PROP. IV.—Theorem. If two triangles (BAC, EDF) have two sides (BA, AC) of one equal respectively to two sides (ED, DF) of the other, and have also the angles (A, D) included by those sides equal, the triangles shall be equal in every respect—that is, their bases or third sides (BC, EF) shall be equal, and the angles (B, C) at the base of one shall be respectively equal to the angles (E, F) at the base of the other; namely, those shall be equal to which the equal sides are opposite.