Plane and solid geometry . GH,A ACD ^ A FHI, A ADE ^ A FIJ. To prove polygon ABODE ^ polygon FGHIJ. Argument 1. In A ABO and FGH, =Z G. 2. AlsoZl = Z2. 3. In A AOD and FHI, Z 3 = Z4. 4. .-. Z1 + Z3=.Z2 + Z4. 5. .-. Z BOD = Z GHI. 6. Likewise Z ODE = Z HIJ, Ze = Zj, and Z ^^5 = Z JFG. 7. .-. polygons ABODE and FGHIJ are mutually equiangular. 8. In A^^aandi^G^,— = — = —. i^G GIT ^F 9. In A AOD and FHI, OAHF CDHI DAIF 10. AndinA^Z)^andi^/e7, — = —= — IF IJ JF 11. ABFG BO CDHI DE EAJF Gil HI IJ12. .-. polygon ABODE ^ polygon FGHIJ. Reasons 1. § 424, 1. 2. § 424, 1. 3. § 424, 1. 4. §54,2
Plane and solid geometry . GH,A ACD ^ A FHI, A ADE ^ A FIJ. To prove polygon ABODE ^ polygon FGHIJ. Argument 1. In A ABO and FGH, =Z G. 2. AlsoZl = Z2. 3. In A AOD and FHI, Z 3 = Z4. 4. .-. Z1 + Z3=.Z2 + Z4. 5. .-. Z BOD = Z GHI. 6. Likewise Z ODE = Z HIJ, Ze = Zj, and Z ^^5 = Z JFG. 7. .-. polygons ABODE and FGHIJ are mutually equiangular. 8. In A^^aandi^G^,— = — = —. i^G GIT ^F 9. In A AOD and FHI, OAHF CDHI DAIF 10. AndinA^Z)^andi^/e7, — = —= — IF IJ JF 11. ABFG BO CDHI DE EAJF Gil HI IJ12. .-. polygon ABODE ^ polygon FGHIJ. Reasons 1. § 424, 1. 2. § 424, 1. 3. § 424, 1. 4. §54,2. 5. § 309. 6. By steps sim- ilar to 1-5. 7. By proof. 8. § 424, 2. 9. §424,2. 10. § 424, 2. 11. §54,1. 12. § 419. 190 PLANE GEOMETRY 439. Cor. Any two similar polygons inay he divided^into the same niimher of triangles similar each to eachand similarly placed, Propositiox XXIII. Problem 440. Upon a line homologous to a side of a given poly-gon, to construct a polygon similar to the given polygon. K-L. Given polygon AB and line MQ homol. to side AE,To construct^ on MQ^ a polygon ^ polygon AD, I. Construction 1. Draw all possible diagonals from A^ as AC and AB, 2. At M^ beginning with MQ as a side, construct A 7, 8, and9 equal respectively to A 1, 2, and 3. § 125. 3. At Q, with MQ as a side, construct Z10 equal to Z 4, andprolong side QF until it meets ML at P. § 125. 4. At P, with PM as a side, construct Zll equal to , andprolong side PO until it meets MR at 0. § 125. 6. At 0, with OM as a side, construct Z12 equal to Z6, andprolong side ON until it meets MF at N. § MP is the polygon required. II. Proof Argument 1. A ABE ^ A MPQ, A ACB ^ A MOP, andA ^PC ^ A MNO. 2. .. polygon MP ~ polygon AB, Reasons 1. §421. 2. § 438. BOOK m 191 IIL The discussion is left as an exercise for the student. Proposition XXIV. Theorem 441. The perimeters of two similar polygons are to eachotJier as any tico homologous sides.
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Keywords: ., bookcentury1900, bookdecade1910, booksubjectgeometr, bookyear1912
