Elements of Geometry: Being Chiefly a Selection from Playfair's GeometryA. Walker, 1829 - 186 páginas |
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... wherefore read , and For 61 read 6 For 2 read 3 For circles read circle For 2 read 3 At the upper right corner of the fig . write E 152 7,10,13,23 For 2 Sup . read 11 158 13 161 15 163 25 163 4 bot . 169 4 175 3 bot . For axis read axes ...
... wherefore read , and For 61 read 6 For 2 read 3 For circles read circle For 2 read 3 At the upper right corner of the fig . write E 152 7,10,13,23 For 2 Sup . read 11 158 13 161 15 163 25 163 4 bot . 169 4 175 3 bot . For axis read axes ...
Página 22
... wherefore CA , AB , BC are equal to one another ; and they form a triangle ABC ; therefore the triangle ABC is equilateral ; and it is described on the given straight line AB . Which was required to be done . * PROPOSITION II . PROBLEM ...
... wherefore CA , AB , BC are equal to one another ; and they form a triangle ABC ; therefore the triangle ABC is equilateral ; and it is described on the given straight line AB . Which was required to be done . * PROPOSITION II . PROBLEM ...
Página 23
... wherefore AE is equal to C ( 1 Ax . ) . Therefore , from AB , the greater of two straight lines , a part AE has been cut off equal to C , the less . Which was to be done . PROPOSITION IV . THEOREM . IF two triangles have two sides of ...
... wherefore AE is equal to C ( 1 Ax . ) . Therefore , from AB , the greater of two straight lines , a part AE has been cut off equal to C , the less . Which was to be done . PROPOSITION IV . THEOREM . IF two triangles have two sides of ...
Página 24
... wherefore also the point C will coincide with F , because AC is equal to DF . But the point B coincides with E ; wherefore the base BC will coincide with EF , and will be equal to it . Therefore the whole trian- gle ABC will coincide ...
... wherefore also the point C will coincide with F , because AC is equal to DF . But the point B coincides with E ; wherefore the base BC will coincide with EF , and will be equal to it . Therefore the whole trian- gle ABC will coincide ...
Página 27
... ; therefore they are equal to each other ( 1 Ax . ) . B Wherefore the three angles of the trian- gle ABC are equal , that is , the triangle ABC is equiangular , ED . A PROPOSITION VI . THEOREM . * If two angles of GEOMETRY . 27 BOOK I.
... ; therefore they are equal to each other ( 1 Ax . ) . B Wherefore the three angles of the trian- gle ABC are equal , that is , the triangle ABC is equiangular , ED . A PROPOSITION VI . THEOREM . * If two angles of GEOMETRY . 27 BOOK I.
Otras ediciones - Ver todas
Elements of Geometry: Being Chiefly a Selection From Playfair's Geometry John Playfair Sin vista previa disponible - 2023 |
Elements of Geometry: Being Chiefly a Selection From Playfair's Geometry John Playfair Sin vista previa disponible - 2023 |
Términos y frases comunes
ABC is equal ABCD alternate angles angle ABC angle ACB angle BAC angles AGH angles equal base ABC bases and altitudes bisect centre chord circumference cone Consequently cylinder demonstrations described diagonals diameter divided draw equal angles equal arches equal bases equal circles equal to AC equiangular Euclid's Euclid's Elements exterior angle fore four quantities four right angles geometry given point given straight line gles greater Hence homologous sides intersect KLMN Let ABC meet opposite angles opposite side paral parallel lines parallel to CD parallelogram parallelopipeds perp perpendicular plane polygon prism Prop pyramid Q. E. D. COR Q. E. D. PROPOSITION radius rectangle contained right angled triangle Scholium segments semicircle side AC similar similar triangles solid square straight line &c subtended tangent THEOREM triangle ABC vertex wherefore
Pasajes populares
Página 36 - Any two sides of a triangle are together greater than the third side.
Página 80 - If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts.
Página 42 - To draw a straight line through a given point parallel to a given straight line. Let A be the given point, and BC the given straight line...
Página 30 - To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given straight line : it is required to divide it intotwo equal parts.
Página 20 - LET it be granted that a straight line may be drawn from any one point to any other point.
Página 38 - Problem. At a given point in a given straight line to make a rectilineal angle equal to a given rectilineal angle.
Página 113 - Wherefore also the angle BAD is equal to the angle CAD : Therefore the angle BAC is cut into two equal angles by the straight line AD.
Página 24 - DE ; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide...
Página 36 - The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it.