Let the segment AB mark the front of the stage. From point A we draw a line AC like this, that the BAC angle is 53 °, that is, the angle considered most appropriate for a convenient view of the stage. Through the point A we draw a line AD perpendicular to the line AC and through the center M of the segment AB we draw a line MN perpendicular to the line AB. Point O, that is the point of intersection of the lines AD and MN, will be the center of a circle passing through points A and B and tangent to line AC. The ADB arc will define the shape we are looking for, which hall should be given, because indeed all angles, whose vertices lie on this arc and whose arms pass through points A and B, will obviously be equal to the ADB angle, and thus the BAC angle, which is clear from considering the ABD triangle, namely: if line BE is perpendicular to line AD, to ∠ ADB = ∠ ABE = ∠ BAC.