The Synoptical Euclid; Being the First Four Books of Euclid's Elements of Geometry, from the Edition of Dr. Robert Simson ... With ExercisesCharles Henry Law, 1854 - 120 σελίδες |
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Αποτελέσματα 6 - 10 από τα 34.
Σελίδα 34
... whole angle ABD is equal to the whole angle ACD ; and the angle BAC has been shown to be equal to the angle BDC ; therefore the opposite sides and angles of parallelograms are equal to one another . Also , their diameter bisects them ...
... whole angle ABD is equal to the whole angle ACD ; and the angle BAC has been shown to be equal to the angle BDC ; therefore the opposite sides and angles of parallelograms are equal to one another . Also , their diameter bisects them ...
Σελίδα 35
... whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also is equal ( I. 34. ) to DC ; and the two EA , AB , are therefore equal to the two FD , DC , each to each ; and the exterior angle FDC is equal ( I. 29 ...
... whole , or the remainder , AE is equal to the whole , or the remainder , DF ; AB also is equal ( I. 34. ) to DC ; and the two EA , AB , are therefore equal to the two FD , DC , each to each ; and the exterior angle FDC is equal ( I. 29 ...
Σελίδα 40
... whole triangle ABC is equal to the whole ADC ; therefore 5. The remaining complement BK is equal to the remaining complement KD . Wherefore , the complements , & c . Q.E.D. PROP . XLIV . - PROBLEM . To a given 40 EUCLID'S ELEMENTS .
... whole triangle ABC is equal to the whole ADC ; therefore 5. The remaining complement BK is equal to the remaining complement KD . Wherefore , the complements , & c . Q.E.D. PROP . XLIV . - PROBLEM . To a given 40 EUCLID'S ELEMENTS .
Σελίδα 43
... whole rectilineal figure ABCD is equal to the whole parallelogram KFLM . Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to be ...
... whole rectilineal figure ABCD is equal to the whole parallelogram KFLM . Therefore the parallelogram KFLM has been described equal to the given rectilineal figure ABCD , having the angle FKM equal to the given angle E. Which was to be ...
Σελίδα 44
... whole angle DBA is equal to the whole FBC . And because the two sides AB , BD , are 44 EUCLID'S ELEMENTS .
... whole angle DBA is equal to the whole FBC . And because the two sides AB , BD , are 44 EUCLID'S ELEMENTS .
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AB is equal AC is equal adjacent angles angle ABC angle ACB angle AGH angle BAC angle BCD angle EDF angle equal base BC bisected circle ABC circumference diameter double draw equal angles equal Constr equal Hyp equal straight lines equal to BC equilateral and equiangular EUCLID'S ELEMENTS exterior angle given circle given point given rectilineal angle given straight line given triangle gnomon greater inscribed interior and opposite less Let ABC Let the straight likewise opposite angles parallel to CD parallelogram pentagon perpendicular point F produced Q.E.D. PROP rectangle AE rectangle contained remaining angle required to describe right angles semicircle side BC square of AC straight line AB straight line AC straight line drawn touches the circle triangle ABC twice the rectangle