 | William Chauvenet - 1871 - 380 sider
...circumscribed about the circle and the other isoperimetrical with the circle. ( Galileo's Theorem.') 265. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. 266. If a = the side of a regular pentagon inscribed in a circle whose radius is R, then, _ a = -771/10... | |
 | William Chauvenet - 1872 - 382 sider
...rectangle whose adjacent sides are equal to the sides of the inscribed and circumscribed squares. 265. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. 2G6. If a = the side of a regular pentagon inscribed in a circle whose radius is It, then, 0 = ^1/10... | |
 | George Albert Wentworth - 1888 - 264 sider
...opposite sides, pass through a common point 0 within the triangle, then OA' OB' OC' = l AA' £B' CV 457. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. Loci. 458. Find the locus of a point P whose distances from two given points A and B are in a given... | |
 | George Albert Wentworth - 1895 - 468 sider
...0 within the triangle, then OA' OB' OC' = l AA' BB' CU 457. Two diagonals of a regular pentagon/not drawn from a common vertex, divide each other in extreme and mean ratio. Loci. 458. Find the locus of a point P whose distances from two given points A and B are in a given... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1896 - 276 sider
...angles at the centre subtended by arcs of equal length are to each other inversely as their radii. 313. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. Hint. — Prove the triangles ABC and BCM similar (§ 275). Then prove AM=AB — BC (§ 77), and substitute... | |
 | George Albert Wentworth - 1896 - 296 sider
...OBC+ A OAC+ A 045 44' A OAB = 0ff A 4.BC CC'' OA' ' AA' OB' OC' But A ABC . OA' EB' L CC" Ex. 457. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. Let BE and AC be diagonals of a regular pentagon and intersect atP. To prove AC : PC = PC : AP. B PROOF.... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1896 - 276 sider
...angles at the centre subtended by arcs of equal length are to each other inversely as their radii. 515, Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. Hint.— Prove the triangles ABC and BCM similar (§ 275). Then prove AM = AB = BC (§ 77), and substitute... | |
 | George Washington Hull - 1897 - 408 sider
...The rectangle of the segments of chords passing through a common point is a constant quantity. 613. Two diagonals of a regular pentagon not drawn from...vertex divide each other in extreme and mean ratio. 814. If three circles intersect one another, their common chords will meet in the same point. BOOKS... | |
 | Andrew Wheeler Phillips, Irving Fisher - 1897 - 374 sider
...subtended by arcs of equal length are to each other inversely as their radii. 495. Exercise.—Two diagonals of a regular pentagon, not drawn from a...vertex, divide each other in extreme and mean ratio. Hint.—Prove the triangles ABC and BCM similar (§ 262). Then prove AM=AB = .5C(§ 76), and substitute... | |
 | George Albert Wentworth - 1899 - 272 sider
...parallel to PC, meeting BC at D, the triangle BPD is equivalent to half the triangle ABC. ' Ex. 551. Two diagonals of a regular pentagon, not drawn from...vertex, divide each other in extreme and mean ratio. Ex. 552. If all the diagonals of a regular pentagon are drawn, another regular pentagon is thereby... | |
| |
Two diagonals of a regular pentagon, not drawn from a common vertex, divide each other in extreme and mean ratio. 
