meets the base, the rectangle of the two sides is equivalent to the rectangle of the segments of the base, together with the square of the bisector. 536. COROLLARY II. If BD be a perpendicular, BE is a diameter, for angle BAE is then right; therefore in any triangle the rectangle of two sides is equivalent to the rectangle of the diameter of the circumscribed circle by the perpendicular to the base from the vertex. EXERCISES. 88. Prove that the inverse of 535 does not hold when AB = BC. 89. Prove 535 when it is an exterior angle which is bisected. THEOREM XVI. 537. The rectangle of the diagonals of a quadrilateral inscribed in a circle is equivalent to the sum of the two rectangles of its opposite sides. also HYPOTHESIS. ABCD an inscribed quadrilateral. To each add × FAC. Then, in As ACD and ABF, X DAC = X BAF; X ACD = X ABD, (376. Angles inscribed in the same segment are equal.) Moreover, ADF = X ACB, (376. Angles inscribed in the same segment are equal.) :. AB. CD + BC.DA = AC.BF + AC. FD = AC(BF + FD) = AC. BD. PROBLEM I. 538. To alter a given sect in a given ratio. GIVEN, the ratio as that of sect a to sect b, and given the sect AB. REQUIRED, to find a sect to which AB shall have the same ratio as a to b. CONSTRUCTION. Make any angle C. On one arm cut off CD = AB. On the other arm cut off CF = a, and FG = b. Join DF, and through G draw GH || DF :. AB : DH :: a : b. 539. This is the same as finding a fourth proportional to three given sects. To find a third proportional to a and b, make AB = b in the above construction. 540. Every alteration of a magnitude is an alteration in some ratio. Two or more alterations are jointly equivalent to some one alteration, and then this single alteration which produces the joint effect of two is said to be compounded of those two. The composition of the ratios of a to b and C to D is performed by assuming F, altering it into G, so that F: G :: a : b, then altering G into H, so that G: H:: C: D. The joint effect turns F into H; and the ratio of F to H is the ratio compounded of the two ratios, a : b and C : D. 541. A ratio arising from the composition of two equal ratios is called the Duplicate Ratio of either. THEOREM XVII. 542. Mutually equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides. HYPOTHESIS. In □ AC, ¥ BCD = × HCF of □ CG. CONCLUSION. AC : □ CG = ratio compounded of DC: CF, and BC: CH. PROOF. Place the s so that HC and CB are in one line; then, by 109, DC and CF are in one line. Complete the □ BF. Then and .. □ AC: BF :: DC: CF, BF:CG :: BC: CH, (505. Parallelograms of equal altitude are as their bases.) AC has to CG the ratio compounded of DC: CF and BC : CH. 543. COROLLARY I. Triangles which have one angle of the one equal or supplemental to one angle of the other, being halves of equiangular parallelograms, are to one another in the ratio compounded of the ratios of the sides about those angles. 544. COROLLARY II. Since all rectangles are equiangular parallelograms, therefore the ratio compounded of two ratios between sects is the same as the ratio of the rectangle contained by the antecedents to the rectangle of the consequents. If the ratio compounded of a: a', and b: b', be written ; and the composition a b this corollary proves b α = ď b ab a'b' of ratios obeys the same laws as the multiplication of fractions. a α Thus 2.2=2, and so the duplicate ratio of two sects is the b b same as the ratio of the squares on those sects. It will be seen hereafter that the special case of 542, when the parallelograms are rectangles, is made the foundation of all mensuration of surfaces. EXERCISES. 90. If mutually equiangular parallelograms are equivalent, so are rectangles with the same sides. 91. Equivalent parallelograms having the same sides as equivalent rectangles are mutually equiangular. |
