Problem.
From a given point (C) in a given right line (AB) to draw a right line perpendicular to the given line.
Sol.—In AC take any point D, and make CE equal to CD [iii.]. Upon DE describe an equilateral triangle DFE [i.]. Join CF. Then CF shall be at right angles to AB.
Dem.—The two triangles DCF, ECF have CD equal to CE (const.) and CF common; therefore the two sides CD, CF in one are respectively equal to the two sides CE, CF in the other, and the base DF is equal to the base EF, being the sides of an equilateral triangle (Def. xxi.); therefore [viii.] the angle DCE is equal to the angle ECF, and they are adjacent angles. Therefore (Def. xiii.) each of them is a right angle, and CF is perpendicular to AB at the point C.