If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. A Geometry Reader - Page 184by Julius J. H. Hayn - 1925 - 316 pagesFull view - About this book
| 1906 - 628 pages
...and QB tangents of the circle O; prove RQ minus RA plus QB. 3. If two chords in a circle intersect, the product of the segments of one chord is equal to the product of the segments of the other chord. 4. The side of an equilateral triangle is a. Find the area. 5. In a circle whose radius is 50 in.,... | |
| Michigan Schoolmasters' Club - Education - 1894 - 554 pages
...study of Boyle's Law. Of this class, is the following: "If two chords be drawn through a fixed point within a circle, the product of the segments of one...equal to the product of the segments of the other." The data could be arranged as follows : ab Pass various lines through P and measure accurately the... | |
| Webster Wells - Geometry - 1894 - 400 pages
...Subtracting (2) from (1), PROPOSITION XXXI. THEOBEM. 280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one...equal to the product of the segments of the other. Let AB and A'B' be any two chords passing through the fixed point P within the circle ABB '. To prove... | |
| Webster Wells - Geometry - 1894 - 394 pages
...— ~BC* = 2ABX DE. PROPOSITION XXXI. THEOREM. 280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one chord is equal to the prod net of the segments of the other. Let AB and A'B' be any two chords passing through the fixed... | |
| George D. Pettee - Geometry, Plane - 1896 - 272 pages
...any other and its external segment. Dem. AB x AD = AC x AE [= AF * ] PROPOSITION XXIII 220. Theorem. If two chords intersect within a circle, the product of the segments of one equals the product of the segments of the other. Dem. x = Y A=D &AEC DEB AE:CE=DE: BE AE x BE = CE... | |
| Webster Wells - Geometry - 1898 - 284 pages
...(2) from (1), we have \ PROP. XXVIII. THEOREM. 280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one...equal to the product of the segments of the other. Given AB and A'B' any two chords passing through fixed point P within O AA'B. To Prove AP x BP = A'P... | |
| Webster Wells - Geometry - 1899 - 450 pages
...278) (§ 277) (1) (2) PROP. XXVIII. THEOREM. 280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one...equal to the product of the segments of the other. Given AB and A'B' any two chords passing through fixed point P within O AA'B. To Prove AP x BP = A'P... | |
| Webster Wells - Geometry - 1899 - 424 pages
...Adding (1) and (2), we have PROP. XXVIII. THEOREM. 280. If any two chords be drawn through a fixed point within a circle, the product of the segments of one...equal to the product of the segments of the other. Given AB and A'B' any two chords passing through fixed point P within O AA'B. To Prove AP x BP = A'P... | |
| Edward Brooks - Geometry, Modern - 1901 - 278 pages
...III. 18. Hence, A EBD is similar to A ECA. Th. 20. Whence, AE : DE =• EC : EB. Def. 5. COR. — Tlie product of the segments of one chord is equal to the product of the segments of the other. For from the preceding proportion we have AE x EB = DE x EC. PROPOSITION XXXII. — THEOREM. If from... | |
| Arthur Schultze, Frank Louis Sevenoak - Geometry - 1901 - 394 pages
...from any point in the perpendicular, AD, to the circles are equal. PROPOSITION XXXIII. THEOREM 312. If two chords intersect within a circle, the product of the segments of one is equal to the product of the segments of the other. Hyp. The chords AB and CD meet in E. To prove... | |
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