Page:Euclid's Elements 1714 Barrow translation.djvu/27

Rh

From a point C in a right line given AB to erect a right line CF at right angles.

a Take on either ide of the point given CD = CE. upon the right line DE b erect an equilateral triangle, draw the line FC, and it will be the perpendicular required.

For the triangles DFC, EFC are mutually c equilateral; d therefore the angle DCF = ECF. e therefore FC is perpendicular. Which was to be done.

The practice of this and the following is eaily performed by the help of a quare.

Upon an infinite right line given AB, from a point given that is not in it, to let fall a perpendicular right line CG.

From the center C a decribe a circle cutting the right line given AB in the points E and F. Then biect EF in G, and draw the right line CG, which will be the perpendicular required.

Let the lines CE, CF be drawn. The triangles EGC, FGC are mutually c equilateral, d therefore the angles EGC, FGC are equal, and by e conequence right. e Wherefore GC is a perpendicular. Which was to be done.

When a right line AB tanding upon a right line CD maketh angles ABC, ABD; it maketh either two right angles, or two angles equal to two right.