16. Proposition thirty-fourth. constructed somewhat differently. This problem is generally In AB take any point C, and on BC (I. 1. cor.) describe DBE are equilateral, the angles CBD and DBE are each of them equal to two-third parts of a right angle (I. 30. cor. 1.); and the triangles BDF, BEF, having the sides BD, DF equal to BE, EF, and the side BF common, are (I. 2.) equal, and consequently the angles FBD and FBE are equal, and each of them the half of DBE. The angle FBD, being therefore one-third part of a right angle, and the angle DBA two-third parts, the whole angle FBC must be an entire right angle, or the straight line BF is perpendicular to AB. BOOK II. 1. A simple proposition might be here introduced. A straight line bisecting two sides of a triangle, is parallel to the base. The straight line DE which joins the middle points of the sides AB and BC, is parallel to the base AC of the triangle ABC. For join AE and CD. Because the triangles ADC, BCD stand on equal bases AD, DB, and have the same vertex or altitude, they are (II. 2.) equivalent, and therefore ADC is half of the whole triangle ABC. For the E the same base AC, and have therefore the same altitude (II. 3.), or DE is parallel to AC. Cor. Hence the triangle DBE cut off by the line DE, is the fourth part of the original triangle. For bisect AC in G, and join DG, which is therefore parallel to BC. The triangle ADG is equivalent to GDC (II. 2.), and GDC, being the half of the rhomboid CE, is equivalent to DEC, which again is (II. 2.) equivalent to DEB. The triangle ABC is thus divided into four equivalent triangles, of which DBE is one. also the rhomboid GDEC is half of the original triangle. Hence 2. From the preceding proposition the following theorem is easily derived: Straight lines joining the successive middle points of the sides of a quadrilateral figure, form a rhomboid. If the sides of the quadrilateral figure ABCD be bisected, and the points of section joined in their order; EFGH is a rhomboid. For draw AC, BD. And because FG bisects AB, BC, it is parallel to AC; and for the same reason, EH, as it bisects AD and DC, is parallel to AC. Wherefore FG is parallel to EH (I. 28.). In like manner, it is proved that EF is parallel to HG; and consequently the figure EFGH is a rhomboid or parallelogram. F B G K H E It is likewise evident, that the inscribed rhomboid is half of the quadrilateral figure; for IG is half of the triangle ABC and IH is half of the triangle ADC. 3. Proposition fourth. This problem is of great use in practical geometry. The plan, for instance, of any grounds, however irregular, is divided into a number of triangles, which are successively reduced to a simple triangle, and this again is converted (by II. 6.) into a rectangle. Instead of computing, therefore, each component triangle, it may be sufficient to calculate the area of the final triangle or rectangle. 4. Proposition ninth. On this proposition is founded the method of offsets, which enters so largely into the practice of land-surveying. In measuring a field of a very irregular shape, the principal points only are connected by straight lines forming sides of the component triangles, and the distance of each remarkable flexure of the extreme boundary is taken from these rectilineal traces. The exterior border of the polygon is therefore considered as a collection of trapezoids, which are measured by multiplying the mean of each pair of offsets or perpendiculars into their base or intermediate distance. 5. Proposition tenth. This beautiful property is easily derived from Propositions fifteenth and sixteenth of Book II. 1. Let ABC be a triangle right-angled at B; produce the base AB till AD be equal to the perpendicular BC; on the compound line BD describe the square BDEF, and make DG and EH equal to AB, and join AG, GH and HC. E The triangles ABC and GDA, having the sides AB, BC evidently equal to DG, AD, and the right angle at B equal to that at D, are (I. 3.) equal. In the same manner, the triangles HEG and CHF are proved to be equal to ABC. But (I. 30.) the exterior angle GAB is equal to the interior angles ADG and AGD, from which take away the equal angles CAB and AGD, and there remains GAC equal to ADG, and consequently a right angle. Wherefore the quadrilateral figure AGHC, having D B C likewise all its sides equal, is a square. But by Prop. 15. Book II. the square BDEF, described on the sum of the sides AB and BC, is equivalent to the squares of those sides, together with twice their rectangle. Now (cor. 5. Book II.) the rectangle under AB and BC is double of the triangle ABC; and consequently the square BDEF is equivalent to the squares of AB and BC, and the four triangles CBA, ADG, GEH and HFC; but the same square is equivalent to the interior square AGHC, with those four triangles; wherefore the squares of the base AB and of the perpendicular BC, are equivalent to the single square described on the hypotenuse AC. 2. From the base AB, cut off a part AD equal to the perpendicular BC, and on the remaining portion BD construct the square BDEF; produce DE and EF, till EG and FH be equal to AD, and join AG, GH, and HC. The triangles CBA, ADG, GEH, and HFC are proved to be equal as before. Again, the angle CAG being equal to the angles CAB and DAG or BCA, the acute angles of the right-angled triangle ABC, is consequently a right angle: Wherefore the quadrilateral figure ACHG is a square. But, by Prop. 16. Book II. the square BDEF, described on BD the difference between the base AB and BC the perpendicular is equivalent, to the squares of AB and BC, diminished by twice their rectangle, or by the D B E H F four triangles CBA, ADG, GEH, and HFC. But the square BDEF is evidently equivalent to the square ACHG described on the hypotenuse AC, diminished by those triangles, and therefore equivalent to the squares of the base AB and of the perpendicular BC. This famous proposition appears to have been brought from the East by Pythagoras. Both the demonstrations now given are common in Persia and India. The second mode, however, 1 would seem to be the favourite, since the figure used is in Hindustan styled the bridal chair or couch, in allusion, no doubt, to its prolific virtues. This figure, and the preceding one, are well adapted for exhibiting the result, by the dissection and transposition of their several parts. The very meagre treatises of geometry written in the ancient Sanscrit language, and the versions of Euclid's Elements by Persian or Arabian commentators, display some variety of such dissections. The method generally adopted is ascribed to the Persian astronomer Nassir Eddin, who flourished in the thirteenth century of our æra, under the munificent patronage of the conqueror Zingis Khan. It may gratify the young student in geometry to see the mode of performing this dissection. Having drawn AL parallel to BF, and IC and GO parallel to DB, place the triangle CKA on CFH, invert the triangle GOA on ADG, place the triangle GOM on AKN, and transfer the small triangle GIN to HLM. In this way, the square AGHC is transformed into the two squares CKLF and ADIK. By reversing the process, the squares of the sides of the rightangled triangle may be compounded into the single square of the hypotenuse. E M G 10 N K D A B 6. It was a favourite speculation with the Greek geometers, to express numerically the sides of a right-angled triangle. The rules which they delivered for that purpose are equally simple and ingenious. For the sake of conciseness, it will be convenient, however, to adopt the language of symbols. Let n denote any odd number; then, -1 and n2+1, will repre according to Plato, 2 n, n2. sent the perpendicular, the base, and hypotenuse, of a rightangled triangle. Thus, n being supposed equal to 3, the numbers thence resulting are 3, 4, and 5, or 6, 8, and 10. These X |