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be equal to the angle GAB, and the angle ABC to the angle ABG (Prop. 4.) .

But the triangles AGB, DFE, are identical; confequently the angles of the triangle DFE will, alfo, be equal to the correfponding angles of the triangle ACB. Q. E. D.

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Let ABC, DEF be each of them right angles; then will ABC be equal to DEF.

For conceive the angle DEF to be applied to the angle ABC, so that the point E may coincide with the point в, and the line ED with the line BA.

And if EF does not coincide with BC, let it fall, if poffible, without the angle ABC, in the direction BG; and produce AB to H.

Then, because the angles ABC, ABG are right angles (by Hyp.), the lines CB, GB will be each perpendicular to AH (Def. 8. 9.)

And, fince a right line which is perpendicular to another right line, makes the angles on each fide of it equal (Def. 8.), the angle CBA will be equal to the angle CBн, and the angle GBA to the angle GBH.

But the angle GBA is greater than the angle CBA, or its equal CBH; confequently the angle GBH is alfo greater

than the angle CBH; that is, a part is greater than the whole, which is abfurd.

The line EF, therefore, does not fall without the angle ABC; and in the fame manner it may be fhewn that it does not fall within it; confequently EF and BC will coincide, and the angle DEF be equal to the angle ABC, as was to be fhewn.

PROP. IX. PROBLEM.

To bifect a given rectilineal angle, that is, to divide it into two equal parts.

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Let BAC be the given rectilineal angle; it is required to divide it into two equal parts.

Take any point D in AB, and from AC cut off AE equal to AD (Prop. 3.), and join DE.

Upon DE describe the equilateral triangle DFE (Prop. 1.), and join AF; then will AF bifect the angle BAC, as was required.

For AD is equal to AE, by conftruction; DF is alfo equal to FE (Def. 16.), and AF is common to each of the triangles AFD, AFE.

But when the three fides of one triangle are equal to the three fides of another, each to each, the angles which are oppofite to the equal fides are, alfo, equal (Prop. 7.)

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The fide DF, 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.

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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 вC (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 triangie, are equal to two fides and the included angle of another, each to each, their bafes will also be equal (Prop. 4.)

The bafe 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.

PROP. XI. PROBLEM.

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

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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.

SCHOLIUM. If the given point be at, or near, the end of AB, the line must be produced.

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PROP. XII. PROBLEM.

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

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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 shall be perpendicular to AB.

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

Join GC, and from the point G, with the distance GC, describe 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. 13.) 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

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