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" Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of' the base, equal to one another, and likewise those which are terminated in the other extremity. "
First principles of Euclid: an introduction to the study of the first book ... - Σελίδα 94
των T S. Taylor - 1880
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...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise...

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...subtend, or arc. opposite to» the equal angles, shall be equal to one another. Prop. VII. Theor. Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise...

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...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise...

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...which the vertex of one triangle is upon a side of the other, needs no demonstration.. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise...

Elements of Geometry: Containing the First Six Books of Euclid: With a ...

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...on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are teminated in the other extremity equal to one another Q,ED PROP. VIlI. THEOR. If two triangles...

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Euclid's Elements of Geometry: The Six First Books. To which are Added ...

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...it are equal, and therefore the sides opposite to them. PROP. VII. THEOR. Upon the same base (AB), and on the same side of it, there cannot be two triangles (ACB, ADB), whose conterminous sides are equal, (namely AC to AD, and BC to BD). For, if possible,...

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...which the vertex of one triangle is upon a side of the other, needs uo demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles, that have their sides which are terminated >n one extremity of the base equal to one another, and likewise...

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...which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have thtir sides, which are terminated in one extremity of the base, equal to another, and likewise...




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