368. Any homogeneous identity of the second degree can be demonstrated geometrically by showing that the areas which represent the two members of the identity are equal. Ex. 1145. Q. E. D. b Prove geometrically the following algebraic formulæ : (a) a(b+c+ d) = ab + ac + bd. *The student is advised in the following exercises to shade the area w represents the left member. PROPOSITION VIII. PROBLEM 369. To transform a rectangle into a square. A H Given rectangle ABCD. Required a square = ABCD. Construction. Construct a mean proportional AF between AB and AD. The square on AF is the required square. 370. REMARK. The mean proportional may be constructed by any of the methods of Book III. The two indicated in the diagram, however, are the most useful ones. Ex. 1146. To transform a parallelogram into a square. Ex. 1149. To transform a pentagon into a square. Ex. 1150. To construct a square equivalent to one third of a given triangle. PROPOSITION IX. PROBLEM 371. To construct a square that shall be any given part of a given square. Required a square equivalent to of 4C. Construction. On AD, lay off AE Draw EF AD, and transform rectangle AEFB into a 372. REMARK. If, instead of 4, the ratio m and n are two given lines, make AE the fourth proportional Ex. 1151. To construct a square three times as large as a given square. Ex. 1152. Given a line AB. To construct another line whose square equals of the square of AB. Ex. 1153. Given a line AB. To construct another line whose square equals twice the square of AB. Ex. 1154. To divide a given square ABCD into three equivalent parts by constructing two other squares which contain angle A. D PROPOSITION X. THEOREM 373. In a right triangle the sum of the squares on the arms is equivalent to the square on the hypotenuse.* H B F Given AH and BF squares on the arms, AD the square on the hypotenuse, of the right triangle ABC. Proof. From B draw BL || AE, intersecting AC and ED in I and I respectively. or Draw BE and KC. ▲ HBC= st. 4, HBC is a straight line. * The shading in this figure is employed merely to point out the two congruent triangles. 374. COR. The square on an arm of a right triangle is equivalent to the difference between the squares on the hypotenuse and the other arm. 375. NOTE. The preceding theorem was first demonstrated by Pythagoras about 550 B.C., and is called after him the Pythagorean Theorem. The proof given here was given by Euclid (about 300 B.C.). Since a square on a line is measured by the square of the line, this theorem may also be demonstrated by the proof in 326, Book III. PROPOSITION XI. PROBLEM 376. To construct a square equivalent to the sum of two given squares. [The solution is left to the student.] |