A Lemma is an auxiliary proposition required in the demonstration of a principal proposition. A Secant or Transversal is a line which cuts a system of lines, a circle, or any other geometrical figure. Congruent figures are those that can be made to coincide by superposition. They agree in shape and size, but differ in position. Hence it follows, by Axiom viii., that corresponding parts or portions of congruent figures are congruent, and that congruent figures are equal in every respect. Rule of Identity.—Under this name the following principle will be sometimes referred to:—“If there is but one X and one Y , then, from the fact that X is Y , it necessarily follows that Y is X.”—Syllabus. PROP. I.—Problem. On a given finite right line (AB) to construct an equilateral triangle. Sol.—With A as centre, and AB as radius, describe the circle BCD (Post. iii.). With B as centre, and BA as radius, describe the circle ACE, cutting the former circle in C. Join CA, CB (Post. i.). Then ABC is the equilateral triangle required. Dem.—Because A is the centre of the circle BCD, AC is equal to AB (Def. xxxii.). Again, because B is the centre of the circle ACE, BC is equal to BA. Hence we have proved. AC = AB, and BC = AB. But things which are equal to the same are equal to one another (Axiom i.); therefore AC is equal to BC; therefore the three lines AB, BC, CA are equal to one another. Hence the triangle ABC is equilateral (Def. xxi.); and it is described on the given line AB, which was required to be done. Questions for Examination. 1. What is the datum in this proposition? 2. What is the quaesitum? 3. What is a finite right line? 4. What is the opposite of finite? 5. In what part of the construction is the third postulate quoted? and for what purpose? Where is the first postulate quoted? 6. Where is the first axiom quoted? 7. What use is made of the definition of a circle? What is a circle? 8. What is an equilateral triangle? Exercises. The following exercises are to be solved when the pupil has mastered the First Book:— 1. If the lines AF, BF be joined, the figure ACBF is a lozenge. 2. If AB be produced to D and E, the triangles CDF and CEF are equilateral. 3. If CA, CB be produced to meet the circles again in G and H, the points G, F, H are collinear, and the triangle GCH is equilateral. 4. If CF be joined, CF2 = 3AB2. 5. Describe a circle in the space ACB, bounded by the line AB and the two circles. PROP. II.—Problem. From a given point (A) to draw a right line equal to a given finite right line (BC). Sol.—Join AB (Post. i.); on AB describe the equilateral triangle ABD [i.]. With B as centre, and BC as radius, describe the circle ECH (Post iii.). Produce DB to meet the circle ECH in E (Post. ii.). With D as centre, and DE as radius, describe the circle EFG (Post. iii.). Produce DA to meet this circle in F. AF is equal to BC.