Because Then in the triangles ABD, FBC, and BD is equal to BC, also the angle ABD is equal to the angle FBC; therefore the triangle ABD is equal to the triangle FBC. 1.4. Now the parallelogram BL is double of the triangle ABD, for they are on the same base BD, and between the same parallels BD, AL. I. 41. And the square GB is double of the triangle FBC, for they are on the same base FB, and between the same parallels FB, GC. But doubles of equals are equal : I. 41. Ax. 6. therefore the parallelogram BL is equal to the square GB. In a similar way, by joining AE, BK, it can be shewn that the parallelogram CL is equal to the square CH. Therefore the whole square BE is equal to the sum of the squares GB, HC: that is, the square described on the hypotenuse BC is equal to the sum of the squares described on the two sides BA, AC. Q.E.D. NOTE. It is not necessary to the proof of this Proposition that the three squares should be described external to the triangle ABC; and since each square may be drawn either towards or away from the triangle, it may be shewn that there are 2×2×2, or eight, possible constructions. EXERCISES. 1. In the figure of this Proposition, shew that (i) If BG, CH are joined, these straight lines are parallel; (ii) The points F, A, K are in one straight line; (iii) FC and AD are at right angles to one another; 2. On the sides AB, AC of any triangle ABC, squares ABFG, ACKH are described both toward the triangle, or both on the side remote from it: shew that the straight lines BH and CG are equal. 3. On the sides of any triangle ABC, equilateral triangles BCX, CAY, ABZ are described, all externally, or all towards the triangle: shew that AX, BY, CZ are all equal. 4. The square described on the diagonal of a given square, is double of the given square. 5. ABC is an equilateral triangle, and AX is the perpendicular drawn from A to BC: shew that the square on AX is three times the square on BX. 6. Describe a square equal to the sum of two given squares. 7. From the vertex A of a triangle ABC, AX is drawn perpendicular to the base: shew that the difference of the squares on the sides AB and AC, is equal to the difference of the squares on BX and CX, the segments of the base. 8. If from any point O within a triangle ABC, perpendiculars OX, OY, OZ are drawn to the sides BC, CA, AB respectively; shew that the sum of the squares on the segments AZ, BX, CY is equal to the sum of the squares on the segments AY, CX, BZ. Let CAB be a right-angled triangle, having the angle at A a right angle: then shall the square on the hypotenuse BC be equal to the sum of the squares on BA, AC. On AB describe the square ABFG. I. 46. From FG and GA cut off respectively FD and GK, each equal to AC. On GK describe the square GKEH: I. 3. I. 46. I. 14. It will first be shewn that the figure CEDB is the square on CB. Now CA is equal to KG; add to each AK: Similarly DH is equal to GF: hence the four lines BA, CK, DH, BF are all equal. Because Then in the triangles BAC, CKE, BA is equal to CK, and AC is equal to KE; Proved. Constr. also the contained angle BAC is equal to the contained angle CKE, being right angles; therefore the triangles BAC, CKE are equal in all respects. 1. 4. Similarly the four triangles BAC, CKE, DHE, BFD may be shewn to be equal in all respects. Therefore the four straight lines BC, CE, ED, DB are all equal; that is, the figure CEDB is equilateral. Again the angle CBA is equal to the angle DBF; then the angle CBD is equal to the angle ABF: Proved. Def. 28. And EHGK is equal to the square on AC. Constr. Now the square on CEDB is made up of the two triangles BAC, CKE, and the rectilineal figure AKEDB; therefore the square CEDB is equal to the triangles EHD, DFB together with the same rectilineal figure; but these make up the squares EHGK, AGFB: hence the square CEDB is equal to the sum of the squares EHGK, AGFB: that is, the square on the hypotenuse BC is equal to the sum of the squares on the two sides CA, AB. Q. E. D. Obs. The following properties of a square, though not formally enunciated by Euclid, are employed in subsequent proofs. [See 1. 48.] (i) The squares on equal straight lines are equal. PROPOSITION 48. THEOREM. If the square described on one side of a triangle be equal to the sum of the squares described on the other two sides, then the angle contained by these two sides shall be a right angle. A Let ABC be a triangle; and let the square described on BC be equal to the sum of the squares described on BA, AC: then shall the angle BAC be a right angle. Construction. From A draw AD at right angles to AC; 1.11. and make AD equal to AB. Join DC. Proof. Then, because AD is equal to AB, 1. 3. Constr. therefore the square on AD is equal to the square on AB. To each of these add the square on CA; then the sum of the squares on CA, AD is equal to the sum of the squares on CA, AB. But, because the angle DAC is a right angle, Constr. therefore the square on DC is equal to the sum of the squares on CA, AD. I. 47. And, by hypothesis, the square on BC is equal to the sum of the squares on CA, AB; therefore the square on DC is equal to the square on BC: therefore also the side DC is equal to the side BC. Then in the triangles DAC, BAC, Because DA is equal to BA, and AC is common to both; Constr. also the third side DC is equal to the third side BC; Proved. therefore the angle DAC is equal to the angle BAC. I. 8. But DAC is a right angle; therefore also BAC is a right angle. Constr. Q. E. D. THEOREMS AND EXAMPLES ON BOOK I. INTRODUCTORY. HINTS TOWARDS THE SOLUTION OF GEOMETRICAL EXERCISES. ANALYSIS. SYNTHESIS. It is commonly found that exercises in Pure Geometry present to a beginner far more difficulty than examples in any other branch of Elementary Mathematics. This seems to be due to the following causes. (i) The main Propositions in the text of Euclid must be not merely understood, but thoroughly digested, before the exercises depending upon them can be successfully attempted. (ii) The variety of such exercises is practically unlimited; and it is impossible to lay down for their treatment any definite methods, such as the student has been accustomed to find in the rules of Elementary Arithmetic and Algebra. (iii) The arrangement of Euclid's Propositions, though perhaps the most convincing of all forms of argument, affords in most cases little clue as to the way in which the proof or construction was discovered. Euclid's propositions are arranged synthetically that is to say, they start from the hypothesis or data; they next proceed to a construction in accordance with postulates, and problems already solved; then by successive steps based on known theorems, they finally establish the result indicated by the enunciation. Thus Geometrical Synthesis is a building up of known results, in order to obtain a new result. But as this is not the way in which constructions or proofs are usually discovered, we draw the attention of the student to the following hints. Begin by assuming the result it is desired to establish; then by working backwards, trace the consequences of the assumption, and try to ascertain its dependence on some simpler theorem which is already known to be true, or on some condition which suggests the necessary construction. If this attempt is successful, the steps of the argument may in general be re-arranged in reverse order, and the construction and proof presented in a synthetic form. |