## Euclid, Elements of Geometry, Book I, Proposition 12 (Edited by Dionysius Lardner, 1855)

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

 (73) To draw a perpendicular to a given indefinite right line (A B). from a point (C) given without it.

### Solution.

Take any point X on the other side of the given line, and from the centre C with the radius C X describe a circle cutting the given line in E and F. Bisect E F in D (X), and draw from the given point to the point of bisection the right line C D; this line is the required perpendicular.

### Demonstration.

For draw C E and C F, and in the triangles E D C and F D C the sides E C and F C, and E D and F D, are equal (const.) and C D common; therefore (VIII) the angles E D C and F D C opposite to the equal sides E C and F C are equal, and therefore D C is perpendicular to the line A B (11).

In this proposition it is necessary that the right line A B be indefinite in length, for otherwise it might happen that the circle described with the centre C and the radius C X might not intersect it in two points, which is essential to the solution of the problem.

It is assumed in the solution of this problem, that the circle will intersect the right line in two points. The centre of the circle being on one side of the given right line, and a part of the circumference (X) on the other, it is not difficult to perceive that a part of the circumference must also be also on the same side of the given line with the centre, and since the circle is a continued line it must cross the right line twice. The properties of the circle form the subject of the third book, and those which are assumed here will be established in that part of the Elements.

The following questions will afford the student useful exercise in the application of the geometrical principles which have been established in the last twelve propositions.

 (74) In an isosceles triangle the right line which bisects the vertical angle also bisects the base, and is perpendicular to the base.

For the two triangles into which it divides the isosceles, there are two sides (those of the isosceles) equal, and a side (the bisector) common, and the angles included by those sides equal, being the parts of the bisected angle; hence (IV) the remaining sides and angles are respectively equal; that is, the parts into which the base is divided by the bisector are equal, and the angle which the bisector makes with the base are equal. Therefore it bisects the base, and is perpendicular to it.

It is clear that the isosceles triangle itself is bisected by the bisector of its vertical angle, since the two triangles are equal.

 (75) It follows also, that in an isosceles triangle the line which is drawn from the vertex to the middle point of the base bisects the vertical angle, and is perpendicular to the line.

For in this case the triangle is divided into two triangles, which have their three sides respectively equal each to each, and the property is established by (VIII)

 (76) If in a triangle the perpendicular from the vertex on the base bisect the base, the triangle is isosceles.

For in this case in the two triangles into which the whole is divided by the perpendicular, there are two sides (the parts of the base) equal, one side (the perpendicular) common, and the included angles equal, being right. Hence (IV) the sides of the triangle are equal.

 (77) To find a point which is equidistance from the three vertical points of a triangle A B C.

Bisect the sides A B and B C at D and E (X), through the points D and E draw perpendiculars, and produce them until they meet at F. The point F is at equal distances from A, B and C.

For draw F A, F B, F C. B F A is isosceles by (76), and for the same reason B F C is isosceles. Hence it is evident that F A, F C, and F B are equal.

 (78) Cor.—Hence F is the centre, and F A the radius of a circle circumscribed about the triangle.
 (79) In a quadrilateral formed by two isosceles triangles A C B and A D B constructed on different sides of the same base, the diagonals intersect at right angles, and that which is the common base of the isosceles triangles is bisected by the other.

For in the triangles C A D and C B D the three sides are equal each to each, and therefore (VIII) the angles A C E and B C E are equal. The truth of the proposition therefore follows from (74)

 (80) Hence it follows that the diagonals of a lozenge bisect each other at right angles.
 (81) It follows from (76) that if the diagonals of a quadrilateral bisect each other at right angles it is a lozenge.

Next: Proposition 13

Previous: Proposition 11

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