In a similar manner it may be shown that AB: A'B' = BC: B'C', .. AABC and A'B'C' have their homologous sides portional, and are similar. 291. COR. 1. Two triangles are similar if two angles one are respectively equal to two angles of the other. 292. COR. 2. Two right triangles are similar if an angle of the one is equal to an acute angle of the other. Ex. 566. Two isosceles triangles are similar if an angle of the equal to the homologous angle of the other. Ex. 567. If two chords AB and CD intersect in E, the triangle is similar to the triangle BED. Ex. 568. If from a point A, without a circle, two secants are dra meet the circumference in B and C, and D and E respectively triangle ABE is similar to the triangle ACD. Ex. 569. The sides of a polygon are 3, 4, 5, 6, and 7. Find the of a similar polygon if the side corresponding with 3 equals 15. Ex. 570. The sides of a triangle are a, b, and c. Find the side: similar triangle if the side corresponding with a is equal to m. Ex. 571. If the altitudes AD and BE of the triangle ABC inters F, the triangle AFE is similar to the triangle BFD. Ex. 572. If the bisector AD of the inscribed triangle ABC be duced to meet the circumference in E, the triangle ABD is similar triangle AEC. 293. REMARK. Similar triangles are the usual mear proving that lines are proportional. To prove, therefore, four lines are proportional. (1) Select two triangles so that each contains two of the lines. (It is advisable to mark the lines as indicated in Ex. (2) Prove the similarity of the two triangles. (If triangle not similar select another pair.) (3) Derive the proportion. Ex. 573. If in the triangle ABC the altitudes AD and BE be drawn, prove that AC: BC DC: EC. = Ex. 574. In the same diagram, if AD and BE meet in F, prove BF: FA= DF: FE. Ex. 575. If from the vertex A of an inscribed triangle ABC the altitude AD and the diameter AF be drawn, then AB: AD = AF: AC. Ex. 576. If from a point without a circle a tangent and a secant be drawn, the tangent is the mean proportional between the secant and its exterior segment. Ex. 577. If a diameter AB be produced to C, at C a perpendicular be erected, and through B a line be drawn to meet the circumference and the perpendicular in D and E, respectively, then AB : BE = DB : BC. Ex. 578. In similar triangles homologous bisectors have the same ratio as any two homologous sides. Ex. 579. In similar triangles homologous altitudes have the same ratio as any two homologous sides. Ex. 580. If two circles are tangent internally, chords of the greater circle drawn from the point of contact are divided proportionally. Ex. 581. The diagonals of a trapezoid divide each other proportionally. Ex. 582. If in a right triangle ABC the altitude AD be drawn upon the hypotenuse, AD: AB = AC: BC. DC: AC. Ex. 583. In the same diagram, AD: AB: Ex. 584. In similar triangles the radii of the inscribed circles have the same ratio as any two homologous sides. 294. REMARK. To prove that the product of two lines equals the product of two other lines, use the method of (293), and take the products of the means and extremes of the resulting proportion. Ex. 585. If two chords intersect within a circumference, the product of the segments of one is equal to the product of the segments of the other. Ex. 586. If from any point E in the chord AB the perpendicular EC be drawn upon the diameter AD, then Ex. 587. In the diagram for Prop. XIII, prove that AD × BC= AB × DE. Ex. 588. The product of two arms of a right triangle is equal to the product of the hypotenuse and the altitude upon the hypotenuse. Ex. 589. The product of any altitude of a triangle and its corresp ing side is equal to the product of any other altitude and its correspon side. Ex. 590. If in the triangle ABC the altitudes AD and BE me F, then BD x DC = DF × AD. Ex. 591. In the same diagram, BD × AC = BF × AD. Ex. 592. If AB is a diameter, BD the tangent at B, and DA meets circumference in E, then AB2 = AE X AD. Ex. 593. In the same diagram, BE2 = AE × ED. Ex. 594. If in an arm AB of an isosceles triangle ABC a point taken so that CD equals the base, CB, then CB2 = BD × BA. PROPOSITION XX. THEOREM 295. Two triangles are similar if an angle of one is equal to an angle of the other, and the sides cluding this angle are proportional. Hyp. In ▲ ABC and A'B'C', To prove ▲ ▲ = ▲ ▲', and AB: A'B' = AC: A'C'. ▲ ABC~A A'B'C'. Place ▲ A'B'C' upon ▲ ABC, so that A' coincides with Ex. 595. Two isosceles triangles are similar if their vertical angles ar equal. Ex. 596. In similar triangles, homologous medians have the same rati as any two homologous sides. Ex. 597. Two triangles, ABC and A'B'C', are similar if altitud AD: altitude A'D' BC: B'C', and Z B = LB'. Ex. 598. In similar triangles, the radii of the circumscribed circles hav the same ratio as any two homologous sides. PROPOSITION XXI. THEOREM 296. Two triangles are similar if their homologous sides are proportional. Proof. AB BC AC = A'B' B'C' A'C' ▲ ABC A A'B'C'. On AC and AB respectively, lay off AD = A'C'' and AE A'B', and draw DE. = Ex. 599. Two triangles are similar if two sides and the median to one of these sides of one are proportional to the homologous parts of the other triangle. * Ex. 600. Two triangles are similar if two sides and the radius of the circumscribed circle of the one are proportional to the homologous parts of the other. *Ex. 601. Two right triangles are similar if the hypotenuse and an arm of one triangle are proportional to the hypotenuse and an arm of the other. PROPOSITION XXII. THEOREM 297. Two triangles are similar if the sides of the one are respectively parallel to the sides of the other. Hyp. The AABC and A'B'C', having AB | A'B', AC || A'C", and BC B'C'. Proof. A and A' are either equal or supplementary. (91) In like manner, ▲ B and B' and C and C" are either equal or supplementary. Hence, one of the following possibilities must be true: (1) The three homologous angles are supplementary, i.e. ZA+ZA'=2 rt. 4, ZB+Z B'=2 rt. 4, ZC+/C=2 rt. . (2) Two angles are supplementary, one is equal to the homologous one, e.g. 2 A + Z A' = 2 rt. 4, ≤ B + ≤ B' = 2 rt. ≤, 2 C = 2 C'. |