3.1 1Ex. 3 occurs in the solution of the problem of the inscription of a regular polygon of seventeen sides in a circle. See note C. Given the difference of two lines = R, and their rectangle = 4R2; find the lines. PROP. IX.—Theorem. If a line (AB) be bisected (at C) and divided into two unequal parts (at D), the sum of the squares on the unequal parts (AD, DB) is double the sum of the squares on half the line (AC), and on the segment (CD) between the points of section. Dem.—Erect CE at right angles to AB, and make it equal to AC or CB. Join AE, EB. Draw DF parallel to CE, and FG parallel to CD. Join AF. Because AC is equal to CE, and the angle ACE is right, the angle CEA is half a right angle. In like manner the angles CEB, CBE are half right angles; therefore the whole angle AEF is right. Again, because GF is parallel to CB, and CE intersects them, the angle EGF is equal to ECB; but ECB is right (const.); therefore EGF is right; and GEF has been proved to be half a right angle; therefore the angle GFE is half a right angle [I. xxxii.]. Therefore [I. vi.] GE is equal to GF. In like manner FD is equal to DB. Again, since AC is equal to CE, AC2 is equal to CE2; but AE2 is equal to AC2 + CE2 [I. xlvii.]. Therefore AE2 is equal to 2AC2. In like manner EF2 is equal to 2GF2 or 2CD2. Therefore AE2 + EF2 is equal to 2AC2 + 2CD2; but AE2 + EF2 is equal to AF2 [I. xlvii.]. Therefore AF2 is equal to 2AC2 + 2CD2. Again, since DF is equal to DB, DF2 is equal to DB2: to each add AD2, and we get AD2 + DF2 equal to AD2 + DB2; but AD2 + DF2 is equal to AF2; therefore AF2 is equal to AD2 + DB2; and we have proved AF2 equal to 2AC2 + 2CD2. Therefore AD2 + DB2 is equal to 2AC2 + 2CD2. Or thus: AD = AC + CD; DB = AC − CD. Square and add, and we getAD2 + DB2 = 2AC2 + 2CD2. Exercises. 1. The sum of the squares on the segments of a line of given length is a minimum when it is bisected. 2. Divide a given line internally, so that the sum of the squares on the parts may be equal to a given square, and state the limitation to its possibility. 3. If a line AB be bisected in C and divided unequally in D, 4. Twice the square on the line joining any point in the hypotenuse of a right-angled isosceles triangle to the vertex is equal to the sum of the squares on the segments of the hypotenuse. 5. If a line be divided into any number of parts, the continued product of all the parts is a maximum, and the sum of their squares is a minimum when all the parts are equal. PROP. X.—Theorem. If a line (AB) be bisected (at C) and divided externally (at D), the sum of the squares on the segments (AD, DB) made by the external point is equal to twice the square on half the line, and twice the square on the segment between the points of section.