PREFATORY NOTE TO THE NEW EDITION. THROUGHOUT this Key references have been altered so as to adapt it for use with the New Edition of our Euclid bearing the date 1900. One other change has been made. A difficulty is often felt-especially in the case of private students-as to the form in which the solution of an elementary geometrical question should be presented; we have therefore given in fuller detail the solutions of the Exercises distributed through the text of Book I. The earlier of these now appear worked out at length. August, 1900. H. S. H. F. H. S. KEY TO EUCLID. BOOK I. EXERCISES. Page 15. 1. Because A is the centre of the circle BCD, ... AFAB. And because B is the centre of the circle CAE, ... BF = AB. [Def. 15.] [Def. 15.] [Ax. 1.] 2. In Prop. 1 it has been shewn that AC and BC are each equal to AB; and in Ex. 1 it has been shewn that AF and BF are each equal to AB. ... AC, BC, AF, BF are equal. Hence the fig. ACBF is a rhombus, since its angles are evidently not right angles. [Def. 32.] 3. With centre B, and radius BA, describe a circle, and produce AB to meet its circumference at C. With centre A, and radius AC, describe a second circle PCQ; and from A draw any st. line to cut the circle PCQ at P. Then AP shall be the required line. Now AC is a diameter, and AB a radius, of the smaller circle; AC twice AB. .. = And APAC, being radii of the circle PCQ; [Def. 15.] 4. Here OD OE, being rådit of the outer circle; [Def. 15.] and OA=OB, being radi of the inner circle : hence the remainder AD-the remainder BE. [Ax. 3.] 5. With centre P and radius PQ, describe a circle cutting AB (produced, if necessary) at X and Y. Then since P is the centre of the circle, ... PQ = PX = PY; [Def. 15.] that is, the distances of the two points X and Y from P are equal to PQ. Here BD = AB, being sides of an equilateral triangle: ... BD BC, a radius of the circle CGH; = .. the point D is on the circumference. 7. On AC describe an equilateral triangle ADC. With centre D, and radius DG, describe the circle GKF. Then it may be shewn as in Prop. 2 that AF = BC. [Ax. 1.] 8. From A and B draw AX and BY each equal to PQ. [1. 2.] With centre A, and radius AX, describe a circle; and with centre B, and radius BY, describe a second circle, cutting the first at C and D. Then either ACB or Join AC, BC and AD, BD. ADB is the triangle required. For AC = AX = PQ; [Def. 15 and Constr.] Note that PQ must be greater than half AB, or the circles will not meet. 9. Draw AX and BY each double of the base AB; and from this point proceed as in the last example. [Ex. 3.] 10. With centre A, and radius AN, describe a circle; and with centre B, and radius BM, describe a second circle, cutting the first at a point C. Join AC, BC. Then in the triangle ABC, AC=AN, and BC= BM. [Def. 15.] 2. (i) In the ▲ LBM, MCN, LB = MC, being halves of equal sides; [Ax. 7.] and BM = CN, for the same reason; = Because and the .. the LBM the ... Because LBM = the LM = MN. MCN, being right angles; [Def. 30.] (ii) In the ▲ ABM, DCM, ABDC, being sides of a square; [Def. 30.] and BM = CM, being halves of equal sides; and the ABM = the DCM, being right angles; [Def. 30.] .. the third side AM the third side DM. = (iii) In the 3 ABM, ADN, Because ABAD, being sides of a square; = and BM DN, being halves of equal sides; and the ABM = the ADN, being right angles; [I. 4.] BC DC, being sides of a square; .. the BCN = the A DCM in all respects; ... BN DM. = [I. 4.] [I. 4.] [L. 4.] 3. In the ▲ BAY, CAX, Because BA=CA, being sides of an isos. triangle; and AY = AX, by hypothesis; and the angle at A is common to both triangles; Because BC=AD, by hypothesis; and DC is common to both; and the BCD = the ADC, by hypothesis; ... BDAC. Page 21. 1. (i) Since AB=AD, being sides of a rhombus, (iii) Hence, adding the equal angles in (i) and (ii), 2. In the ABC, since AB = AC, by hypothesis, [1. 4.] [1. 5.] [Ax. 2.] [1. 5.] [Hyp.] [1. 5.] [Ax. 2.] Hence the whole ABD = the whole ACD. 3. As in the last example it may be shewn that |