## An elementary course of mathematics, Volum 2 |

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ABCD altitude angle contained angle of inclination base centre circle circumscribed coincide common cone consequently construction contained cylinder described determined diameter differ dihedral angle distance draw drawn edges equal faces figure Geometry given angle given plane given point given straight line greater horizontal plane horizontal projection horizontal trace inscribed intersection length less line joining line parallel manner meet opposite parallel parallel to xy parallelepiped parallelogram pendicular perimeter perpendicular perpendicular to xy plane angles plane MN plane passing planes of projection polygon position prism Prob PROBLEM produced Prop proved pyramid ratio rectangle represented respectively right angles scale of slope shown sides similar solid solid angle space square surface THEOR third triangle trihedral true turned vertical plane vertical projection vertical trace Wherefore

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Side 5 - If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, viz.

Side 18 - FD. Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN are cut by the plane EBDX, the common sections EX, BD are parallel (Prop.

Side 13 - For the same reason, CD is likewise at right angles to the plane HGK. Therefore AB, CD are each of them at right angles to the plane HGK.

Side 4 - BC above it : and since the straight line AB is in the plane, it can be produced in that plane : let it be produced to D ; and let any plane pass through the straight line AD, and be turned about it until it pass through the point C; and because the points B, C, are in this plane, the straight line* BC is in it: »7Def.1.

Side 9 - Note. (3. 11.) line; let this be BF: therefore the three straight lines AB, BC, BF are all in one plane, viz. that which passes through AB, BC : and because AB stands at right angles to each of the straight lines BD, BE, it is also at right angles (4. 1 1.) to the plane passing through them; and therefore makes right angles (3.

Side 16 - BGH are together equal* to two right angles: and BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is perpendicular...

Side 9 - If three straight lines meet all in one point, and a straight line stand at right angles to each of them in that point ; these three straight lines are in one and the same plane. Let the straight line AB stand at right angles to each of the straight lines BC, BD, BE, in B, the point where they meet ; BC, BD, BE are in one and the same plane. If not, let...

Side 1 - A plane is perpendicular to a plane, when the straight lines drawn in one of the planes perpendicular to the common section of the two planes, are perpendicular to the other plane. 5. The inclination of a straight line to a plane...

Side 28 - Cor. 1.) therefore all the angles of the triangles are equal to all the angles of the polygon together with four right angles : (i. ax. 1.) but all the angles at the bases of the triangles are greater than all the angles of the polygon, as has been proved ; wherefore the remaining angles of the triangles, viz. those of the vertex, which contain the solid angle at A, are less than four right angles.

Side 5 - If a straight line stand at right angles to each of two straight lines in the point of their intersection, it will also be at right angles to the plane in which these lines are.