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.

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 bifeft 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 BC (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 also be equal (Prop. 4.)

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

་་

PRO P. XI. PROBLEM:

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

D

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 sides 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.

1

C

PROF.

PRO P. XII. PROBLEM.

To draw a right line perpendicular to a given right line, of an unlimited length, from a given point without it.

m

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

Take any point D, in AB, and from that point, with the diftance DC, defcribe the circle CGE, cutting AB

in G.

Join GC, and from the point G, with the distance GC, `defcribe the circle n E m, cutting the former in E.

Through the points C, E draw the right line CFE, cutting AB in F, and CF will be perpendicular to AB, as was required.

For, join the points D, C, D, E, and G, E :

Then, because DC is equal to DE, GC to GE (Def. 18.)

and DG common to each of the triangles DCG, DEG, the angle CDG will be equal to the angle GDE (Prop. 7.)

And, fince DC is equal to DE, DF common to each of the triangles DCF, DEF, and the angle CDG equal

to the angle GDE, the angle DFC will also be equal to the angle DFE (Prop. 4.)

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

PROP. XIII. THEOREM.

The angles which one right line makes with another, on the fame fide of it, are together equal to two right angles.

E

1

Let the right line AB fall upon the right line CD; then will the angles ABC, ABD, taken together, be equal to two right angles.

For if the angles ABC, ABD be equal to each other, they will be, each of them, right angles (Def. 8 and 9.) But if they be unequal, let EB be drawn, from the point B, perpendicular to CD (Prop. 11.)

Then, fince the angles EBC, EBD are right angles (Def. 8.), and the angle EBD is equal to the angles EBA, ABD (Ax. 8.), the angles EBC, EBA and ABD will be equal to two right angles.

But the angles EBC, EBA are, together, equal to the angle ABC (Ax. 8.); confequently the angles ABC, ABD are, also, equal to two right angles. Q. E. D.

COROLL. All the angles which can be made, at any point B, on the fame fide of the right line CD, are, together, equal to two right angles.

If a right line meet two other right lines, in the fame point, and make the angles on each fide of it together equal to two right angles, thofe lines will form one continued right line.

B

Let the right line AB meet the two right lines CB, BD, at the point в, and make the angles ABC, ABD together equal to two right angles, then will BD be in the fame right line with CE.

For, if it be not, let fome other line BE be in the fame right line with CB.

Then, because the right line AB falls upon the right line CBE, the angles ABC, ABE, taken together, are equal to two right angles (Prop. 13.)

But the angles ABC, ABD are also equal to two right angles (by Hyp.); confequently the angles ABC, ABE are equal to the angles ABC, ABD.