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

aa 3. 1.
b 1. 1. Take on either ſide 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 cc conſtr.
d 8. 1.
e 10. def. 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 eaſily performed by the help of a ſquare.

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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 aa 3. poſt.
b 10. 1. deſcribe a circle cutting the right line given AB in the points E and F. Then biſect 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 cc conſtr.
d 8. 1.
e 10. def. equilateral, d therefore the angles EGC, FGC are equal, and by e conſequence right. e Wherefore GC is a perpendicular. Which was to be done.

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