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Βιβλία Βιβλία 1 - 10 από 59 για Consequently, a line drawn from the vertex of an isosceles triangle to the middle....
" Consequently, a line drawn from the vertex of an isosceles triangle to the middle of the base, bisects the vertical angle, and is perpendicular to the base. "
An Introduction to the Theory and Practice of Plane and Spherical ... - Σελίδα 138
των Thomas Keith - 1810 - 420 σελίδες
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Elements of Geometry

Adrien Marie Legendre - 1819 - 208 σελίδες
...DAC, and that the angle BDA — ADC ; therefore these two last are right angles. Hence, a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, anil divides the vertical angle into two equal parts. In a triangle that is not isosceles,...

The new practical builder, and workman's companion, Τόμος 2

Peter Nicholson - 1823
...COROLLARY 1. — Hence every equilateral triangle is also equiangular. 62. COROLLARY 2. — A straight line drawn from the vertex of an isosceles triangle to the middle of the base will bisect the vertical angle, and be perpendicular to the base. THEOREM 12. 63. If two angles of...

Elements of Geometry

Adrien Marie Legendre - 1825 - 224 σελίδες
...BAD = DAC, and that the angle BDA = ADC; therefore these two last are right angles. Hence a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts. . In a triangle that is not isosceles,...

Elements of Geometry...: Translated from the French for the Use of the ...

Adrien Marie Legendre, John Farrar - 1825 - 224 σελίδες
...— DAC, and that the angle BDA = ADC ; therefore these two last are right angles. Hence a straight line drawn from the vertex of an isosceles triangle, to the middle of the base, is perpendicular to that base, and divides the vertical angle into two equal parts. In a triangle that is not isosceles,...

An Introduction to the Theory and Practice of Plain and Spherical ...

Thomas Keith - 1826 - 442 σελίδες
...the triangle ADC is equal to the triangle ВЕС (E. 142.); consequently AE+ EC is equal to ED + DC. viz. equal sides are opposite to equal angles. (H)...perpendicular to the base. For the two sides FB and вс are equal to the two sides FA and AC, and the angle FBC is equal to the angle FAC, therefore the...

Elements of Geometry Upon the Inductive Method

James Hayward - 1829 - 172 σελίδες
...AC ; PB and PC will be equal (41), and AB and AC will therefore be equal ; and AG will be a straight line drawn from the vertex of an isosceles triangle to the middle of the base ; that is — BC is perpendicular to the oblique line AG, when it is perpendicular to the straight...

Elements of Geometry and Trigonometry: With Notes

Adrien Marie Legendre - 1830 - 316 σελίδες
...that the angle BAD is equal to DAC, and BDA to ADC ; hence the latter two are right angles ; hence the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to that base, and divides the angle at the vertex into two equal...

Elements of Geometry and Trigonometry

Adrien Marie Legendre - 1836 - 359 σελίδες
...angle BAD, is equal to DAC, and BDA to ADC, hence the latter two are right angles ; therefore, the line drawn from the vertex of an isosceles triangle to the middle point of its base, is perpendicular to the base, and divides the angle at the vertex into two equal...

An Elementary Treatise on Plane and Solid Geometry

Benjamin Peirce - 1837 - 159 σελίδες
...right angle ; and also DAB = DAC, that . is> The arc, drawn from the vertex of an isosceles spherical triangle to the middle of the base, is perpendicular to the base, and bisects the angle at the vertex. 454. Corollary. An equilateral spherical triangle is also equiangular....

An introduction to the theory ... of plane and spherical trigonometry ...

Thomas Keith - 1839
...(314) COROLLARY II. Hence every equilateral triangle is likewise equiangular, and the contrary. (315) COROLLARY III. A line drawn from the vertex of an...therefore the angle CFB is equal to the angle CFA (310) ; but CFA and CFB are equal to two right angles (294), therefore CF is perpendicular to AB. (316)...




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