PROPOSITIONS 1-26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS. PROPOSITION XI. To draw a straight line at right angles to a given straight line, from a given point in the same. be the given straight line, and a given point in it. Let It is required to draw a straight line from the point at right angles to ides ase ; therefore the angle and join is equal to Then drawn from the point shall be at right angles to and is common to the two triangles ; the two each to each; and the base is equal to the and these two angles are adjacent is equal to the angle ngles. But when the two adjacent angles which one straight line makes with another straight line, are qual to one another, each of them is called a right angle: therefore each of the angles s a right angle. Wherefore from the given point , in the given straight line ngles to COR. By help of this problem, it may be demonstrated that two straight lines cannot have a ommon segment. If it be possible, let the segment is a straight line, therefore is equal to the angle the less equal to the greater angle, which is Therefore two straight lines cannot have a common segment. PROPOSITIONS 1-26, BOOK I, ARE NOW PUBLISHED IN A SIMILAR FORM TO THIS. PROPOSITION X. To bisect a given finite straight line, that is, to divide it into two equal parts. shall be cut into two equal parts in the point |