154. Let ACB, ADB be two right-angled triangles having a common hypotenuse AB, join CD, and on CD produced both ways draw perpendiculars AE, BF. Shew that CE+CF2 = DE2 + DF2. 155. If perpendiculars AD, BE, CF, drawn from the angles on the opposite sides of a triangle intersect in G, the difference of the squares on the sides AC, AB, is equal to the difference of the squares on the lines CG BG. 156. If ABC be a triangle of which the angle A is a right angle; and BE, CF be drawn bisecting the opposite sides respectively: shew that four times the sum of the squares on BE and CF is equal to five times the square on BC. 157. If ABC be an isosceles triangle, and CD be drawn perpendicular to AB; the sum of the squares on the three sides is equal to AD2 +2.BD2 + 3. CD3. 158. The sum of the squares described upon the sides of a rhombus is equal to the squares described on its diameters. 159. A point is taken within a square, and straight lines drawn from it to the angular points of the square, and perpendicular to the sides; the squares on the first are double the sum of the squares on the last. Shew that these sums are least when the point is in the center of the square. 160. In the figure Euc. 1. 47, (a) Shew that the diagonals FA, AK of the squares on AB, AC, lie in the same straight line. (b) If DF, EK be joined, the sum of the angles at the bases of the triangles BFD, CEK is equal to one right angle. If BG and CH be joined, those lines will be parallel. If perpendiculars be let fall from Fand K on BC produced, the parts produced will be equal; and the perpendiculars together will be equal to BC. (e) Join GH, KE, FD, and prove that each of the triangles so formed, equals the given triangle ABC. (ƒ) The sum of the squares on GH, KE, and FD will be equal to six times the square on the hypotenuse. (g) The difference of the squares on AB, AC, is equal to the difference of the squares on AD, AE. 161. The area of any two parallelograms described on the two sides of a triangle, is equal to that of a parallelogram on the base, whose side is equal and parallel to the line drawn from the vertex of the triangle, to the intersection of the two sides of the former parallelograms produced to meet. 162. If one angle of a triangle be a right angle, and another equal to two-thirds of a right angle, prove from the First Book of Euclid, that the equilateral triangle described on the hypotenuse, is equal to the sum of the equilateral triangles described upon the sides which contain the right angle. BOOK II. DEFINITIONS. I. EVERY right-angled parallelogram is called a rectangle, and is said to be contained by any two of the straight lines which contain one of the right angles. II. In every parallelogram, any of the parallelograms about a diameter together with the two complements, is called a gnomon. "Thus the parallelogram HG together with the complements AF, FC, is the gnomon, which is more briefly expressed by the letters AGK, or EHC, which are at the opposite angles of the parallelograms which make the gnomon." PROPOSITION I. THEOREM. If there be two straight lines, one of which is divided into any number of parts; the rectangle contained by the two straight lines, is equal to the rectangles contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC' be divided into any parts BD, DE, EC, in the points D, E. Then the rectangle contained by the straight lines A and BC, shall be equal to the rectangle contained by A and BD, together with that contained by A and DE, and that contained by A and EC. From the point B, draw BF at right angles to BC, (1. 11.) through G draw GH parallel to BC, (1. 31.) and through D, E, C, draw DK, EL, CH parallel to BG, meeting GH in K, L, H. Then the rectangle BH is equal to the rectangles BK, DL, EH. for it is contained by GB, BC, and GB is equal to 4: because DK, that is, BG, (I. 34.) is equal to A; and in like manner the rectangle EH is contained by A, EC: therefore the rectangle contained by A, BC, is equal to the several rectangles contained by A, BD, and by A, DE, and by A, EC. Wherefore, if there be two straight lines, &c. Q. E. D. PROPOSITION II. THEOREM. If a straight line be divided into any two parts, the rectangles contained by the whole and each of the parts, are together equal to the square on the whole line. Let the straight line AB be divided into any two parts in the point C. Then the rectangle contained by AB, BC, together with that contained by AB, AC, shall be equal to the square on AB. Upon AB describe the square ADEB, (1. 46.) and through C draw CF parallel to AD or BE, (1. 31.) meeting DE in F. Then AE is equal to the rectangles AF, CE. and AF is the rectangle contained by BA, AC; for BE is equal to AB: therefore the rectangle contained by AB, AC, together with the rectangle AB, BC is equal to the square on AB. If therefore a straight line, &c. Q.E.D. PROPOSITION III. THEOREM. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square on the aforesaid part. Let the straight line AB be divided into any two parts in the point C. Upon BC describe the square CDEB, (1. 46.) and produce ED to F, for it is contained by AB, BE, of which BE is equal to BC: therefore the rectangle AB, BC, is equal to the rectangle AC, CB, together with the square on BC. If therefore a straight line be divided, &c. Q. E.D. If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts in C. Then the square on AB shall be equal to the squares on AC, and CB, together with twice the rectangle contained by AC, CB. Upon AB describe the square ADEB, (1. 46.) join BD, through C draw CGF parallel to AD or BE, (1. 31.) meeting BD in G and DE in F; and through G draw HGK parallel to AB or DE, meeting AD in H, and BE in K; Then, because CF is parallel to AD and BD falls upon them, therefore the exterior angle BGC is equal to the interior and opposite angle BDA; (1. 29.) but the angle BDA is equal to the angle DBA, (1. 5.) wherefore the angle BGC is equal to the angle DBA or GBC; and therefore the side BC is equal to the side CG; (1. 6.) but BC is equal also to GK, and CG to BK; (1. 34.) wherefore the figure CGKB is equilateral. It is likewise rectangular ; for, since CG is parallel to BK, and BC meets them, therefore the angles KBC, BCG are equal to two right angles; (1. 29.) but the angle KBC is a right angle; (def. 30. constr.) wherefore BCG is a right angle: and therefore also the angles CGK, GKB, opposite to these, are right angles; (1. 34.) wherefore CGKB is rectangular : but it is also equilateral, as was demonstrated; wherefore it is a square, and it is upon the side CB. For the same reason HF is a square, and it is upon the side HG, which is equal to AC. (1.34.) Therefore the figures HF, CK, are the squares on AC, CB. And because the complement AG is equal to the complement GE. (1. 43.) and that AG is the rectangle contained by AC, CB, therefore GE is also equal to the rectangle AC, CB; wherefore the four figures HF, CK, AG, GE, are equal to the squares on AC, CB, and twice the rectangle AC, CB: but HF, CK, AG, GE make up the whole figure ADEB, which is the square on AB; therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle AC, CB. Wherefore, if a straight line be divided, &c Q. E.D. COR. From the demonstration, it is manifest, that the parallelograms about the diameter of a square, are likewise squares. PROPOSITION V. THEOREM. If a straight line be divided into two equal parts, and also into two unequal parts; the rectangle contained by the unequal parts, together with the square on the line between the points of section, is equal to the square on half the line. Let the straight line AB be divided into two equal parts in the point C, and into two unequal parts in the point D. Then the rectangle AD, DB, together with the square on CD, shall be equal to the square on CB. |