The fide Dr, therefore, being equal to the fide FE, the angle DAF will be equal to the angle FAE; and confequently the angle BAC is bifected by the right line af, as was to be done.

PROP. X. PROBLEM.

To bifect a given finite right line, that is, to divide it into two equal parts.

B

D

Let AC be the given right line; it is required to divide it into two equal parts.

Upon AC defcribe the equilateral triangle ACB (Prop. 1.), and bifect the angle ABC by the right line BD (Prop. 9.); then will AC be divided into two equal parts at the point D, as was required.

For AB is equal to вс (Def. 16.), BD is common to each of the triangles ADB, CDB, and the angle ABD is equal to the angle CBD (by Conft.)

But when two fides and the included angle of one triangle, are equal to two fides and the included angle of another, each to each, their bafes will alfo be equal (Prop. 4.)

The base AD is, therefore, equal to the bafe DC; and, confequently, the right line AC is bifected in the point D, as was to be done.

PROP. XI. PROBLEM.

T

At a given point, in a given right line, to erect a perpendicular.

Let AB be the given right line, and D the given point in it; it is required to draw a right line, from the point D, that shall be perpendicular to AB.

Take any point E, in AB, and make DF equal to DE (Prop. 3.), and upon EF defcribe the equilateral triangle ECF (Prop: 1.)

Join the points D, C; and the right line CD will be perpendicular to AB, as was required.

For CE is equal to CF (Def. 16.), ED to DF (by Conft.) and CD is common to each of the triangles ECD, FCD.

The three fides of the triangle ECD being, therefore, equal to the three fides of the triangle FCD, each to each, the angle EDC will, alfo, be equal to the angle FDC (Prop. 7.)

But one right line is perpendicular to another when the angles on both fides of it are equal (Def. 8.); therefore CD is perpendicular to AB; and it is drawn from the point D as was to be done.

SCHOLLUM. If the given point be at, or near, the end of AB, the line muft be produced.

PROP. II. PROBLEM.

From a given point to draw a right line equal to a given finite right line.

B

Let A be the given point, and BC the given right line; it is required to draw a right line from the point A, that fhall be equal to BC.

Join the points A, B, (Pof. 1.); and upon BA defcribe the equilateral triangle BAD (Prop. 1.).

From the point B, at the distance BC, defcribe the circle CEF (Pof. 3.) cutting DB produced in F.

And from the point D, at the distance DF, defcribe the circle FHG (Pof. 3.); then, if DA be produced to Q, AG will be equal to BC, as was required.

For, fince B is the centre of the circle CEF, BC is equal to BF (Def. 13.)

And, because D is the centre of the circle FHG, DG is equal to DF (Def. 13.)

But the part DA is alfo equal to the part DB (Def. 16.), whence the remainder AG will be equal to the remainder BF (Ax. 3.)

And fince AG, BC have been each proved to be equal to BF, AG will alfo be equal to BC (Ax. 1.)

A right line AG, has, therefore, been drawn from the point A, equal to the right line BC, as was to be done.

SCHOLIUM. When the point A is at one of the extremities B, of the given line BC, any right line, drawn from that point to the circumference of the circle CEF, will be the one required.

From the greater of two given right lines, to cut off a part equal to the lefs.

Let AB and c be the two given right lines; it is required to cut off a part from AB, the greater, equal to c the lefs.

From the point A draw the right line AD equal to c (Prop. 2.); and from the centre A, at the distance AD, defcribe the circle DER (Pf. 5.) cutting AB in E, and AE will be equal to C as was required.

For, fince A is the centre of the circle EDF, AE will be equal to AD (Def. 13.)

But c is equal to AD, by conftruction; therefore AE will alfo be equal to c (Ax. 1.)

Whence, from AB, the greater of the two given lines, there has been taken a part equal to c the lefs, which was to be done.

SCHOLIUM. When the two given lines are fo fituate, that one of the extremities of c falls in the point A, the former part of the construction becomes unneceffary.

PROP.

If two fides and the included angle of one triangle, be equal to two fides and the included angle of another, each to each, the triangles will be equal in all refpects.

Let ABC, DEF be two triangles, having CA equal to FD, CB to FE, and the angle c to the angle F; then will the two triangles be equal in all refpects.

For conceive the triangle ABC to be applied to the triangle DEF, fo that the point c may coincide with the point F, and the fide CA with the fide FD.

Then, because CA coincides with FD, and the angle c is equal to the angle F (by Hyp.), the fide CB will also coincide with the fide FE.

And, fince CA is equal to FD, and CB to FE (by Hyp.), the point A will fall upon the point D, and the point B upon the point E.

But right lines, which have the fame extremities, must coincide, or otherwise their parts would not lie in the fame direction, which is abfurd (Def. 5.); therefore AB falls upon, and is equal to DE.

And, because the triangle ABC coincides with the triangle DEF, the angle A will be equal to the angle D, the angle B to the angle E, and the two triangles will be equal in all refpects (Ax. 9.) QE. D.