## Euclid, Elements of Geometry, Book I, Proposition 11 (Edited by Dionysius Lardner, 1855)

Proposition XI. Problem.
[Euclid, ed. Lardner, 1855, on Google Books]

 (72) From a given point (C) in a given right line (A B) to draw a perpendicular to the given line.

### Solution.

In the given line take any point D and make C E equal to C D (III); upon D E describe an equilateral triangle D F E (I); draw F C, and it is perpendicular to the given line.

### Demonstration.

Because the sides D F and D C are equal to the sides E F and E C (const.), and C F is common to the triangles D F C and E F C, therefore (VIII) the angles opposite to the equal sides D F and E F are equal, and therefore F C is perpendicular to the given right line A B at the point C.

Cor.—By help of this problem it may be demonstrated, that two straight lines cannot have a common segment.

It it be possible, let the two straight lines A B C, A B D have the segment A B common to both of them. From the point B draw B E at right angles to A B; and because A B C is a straight line, the angle C B E is equal to the angle E B A; in the same manner, because A B D is a straight line, the angle D B E is equal to the angle E B A; wherefore the angle D B E is equal to the angle C B E, the less to the greater, which is impossible; therefore the two straight lines cannot have a common segment.

If the given point be at the extremity of the given right line, it must be produced, in order to draw the perpendicular by this construction.

In a succeeding article, the student will find a method of drawing a perpendicular through the extremity of a line without producing it.

The corollary to this proposition is useless, and is omitted in some editions.

It is equivalent to proving that a right line cannot be produced through its extremity in more than one direction, or that it has but one production.

Next: Proposition 12

Previous: Proposition 10

This proposition in other editions: