If the perpendiculars from the vertices to the opposite faces of a tetrahedron be concurrent, then a sphere passes through the four feet of the perpendiculars, and consequently through the centre of gravity of each of the four faces, and through the mid-points of the segments of the perpendiculars between the vertices and their common point of intersection. [Please select]
0
On the other perpendiculars, the points for off-centering are laid off, measuring the required distance on both sides of the center point. [Please select]
0
For example, it is known that if we draw perpendiculars to the sides of a triangle from the opposite angles, all three perpendiculars meet in a point. [Please select]
Do you have a better example in your mind? Please submit your sentence!