H

B

If the sides AB and BC of the given angle be supposed equal, only one circle would be required, a series of equal isosceles triangles being constituted about its centre. It is evident that this addition is without limit, and that the angle so produced may continue to spread out, and its opening side even make repeated revolutions.

PROP. V. PROB.

To bisect a given angle.

Let ABC be an angle which it is required to bisect. In the side AB take any point D, and from BC cut off BE equal to BD; join DE, on which construct (I. 1.) the isosceles triangle DFE, and draw the straight line BF: The angle ABC is bisected by BF.

For the two triangles DBF and EBF, having the side DB equal to EB, the side DF to EF, and BF common to both, are (I. 2.) equal,

A

B

D

E

F

and consequently the angle DBF is equal to EBF.

Cor. Hence the mode of drawing a perpendicular from

a given point B in the straight line AC; for the angle ABC, which the opposite seg

ments BA and BC make with each other, being equal to two right angles, the straight line that bisects it must be the perpendicular required. Taking BD, therefore, equal to BE, and

B

constructing the isosceles triangle DFE; the straight line BF, which joins the vertex of the triangle, is perpendicular to AC.

Schol. In the general construction, the isosceles triangle DFE may stand either below or above the base DE; but if it were made equal to DBE, the vertex F would coincide with B, and render the construction indeterminate.

PROP. VI. PROB.

To let fall a perpendicular upon a straight line, from a given point above it.

From the point C, to let fall a perpendicular upon the given straight line AB.

In AB take towards A the point D, and with the distance DC describe a circle; and, in the same line, take towards B another point E, and with the distance EC describe a second circle intersecting the former in F; join CF, crossing the given line in G: CG is perpendicular to AB.

For the straight lines DC, DF and EC, EF being joined, the triangles DCE and DFE have the side

DC equal to DF, EC to EF, and

-B

DE common to them both; whence (I. 2.) the angle CDE or CDG is equal to FDE or FDG. And because, in the triangles DCG and DFG, the side DC is equal to DF, DG common, and the contained angles CDG and FDG are proved to be equal; these subordinate triangles are (I. 3.) equal, and consequently the angle DGC is equal to DGF, and each of them a right angle, or CG is perpendicular to AB.

PROP. VII. PROB.

To bisect a given finite straight line.

On the given straight line AB, construct two isosceles triangles (I. 1.) ACB and ADB, and join their vertices C and D by a straight line cutting AB in the point E: AB is bisected in E.

E

B

For the sides AC and AD of the triangle CAD being respectively equal to BC and BD of the triangle CBD, and the side CD common to them.both; these triangles (I. 2.) are equal, and the angle ACD or ACE is equal to BCD or BCE. Again, the inferior triangles ACE and BCE, having the side AC equal to BC, CE common, and the contained angle ACE equal to BCE, are (I. 3.) equal, and consequently the base AE is equal to BE.

PROP. VIII. THEOR.

The exterior angle of a triangle is greater than either of its interior opposite angles.

The exterior angle BCF, formed by producing a side AC of the triangle ABC, is greater than either of the opposite and interior angles CAB and CBA.

For bisect the side BC in the point D (I. 7.), draw AD, and produce it until DE be equal to AD, and join EC.

G

F

The triangles ADB and EDC have, by construction, the side DA equal to DE, the side DB to DC, and the vertical angle BDA equal to CDE; these triangles are, therefore, equal (I. 3.), and the angle DCE is equal to DBA. But the angle BCF is evidently greater than DCE; it is consequently greater than DBA or CBA. qr

In like manner, it may be shown, that if BC be produced, the exterior angle ACG is greater than CAB. But 'ACG is equal to the vertical angle BCF, and hence BCF must be greater than either the angle CBA or CAB.

Cor. Hence all the exterior angles of a triangle are

greater than the interior, and likewise greater than three right angles.

PROP. IX. THEOR.

Any two angles of a triangle are together less than two right angles.

The two angles BAC and BCA of the triangle ABC are together less than two right angles.

For produce the common side AC. And, by the last proposition, the exterior angle BCD is greater than BAC, add BCA to each, and the

two angles BCD and BCA are greater than BAC and BCA, or BAC and BCA are together less than BCD and BCA, that is, less than two right angles (Def. 4).

Cor. Hence a triangle can only have one right or obtuse angle, its two remaining angles being always acute.

PROP. X. THEOR.

The angles at the base of an isosceles triangle are equal.

The angles BAC and BCA at the base of the isosceles triangle ABC are equal.

For draw (I. 5.) BD bisecting the vertical angle ABC. Because, by hypothesis, AB is equal to

BC, the side BD common to the two triangles BDA and BDC, and the angles ABD and CBD contained by them are

equal; these triangles are equal (I. 3.), and consequently the angle BAD is equal to BCD.

B

D

Cor. Every equilateral triangle is also equiangular.

PROP. XI. THEOR.

If two angles of a triangle be equal, the sides opposite to them are likewise equal.

Let the triangle ABC have two equal angles BCA and BAC; the opposite sides AB and BC are also equal.

For if AB be not equal to CB, let it be equal to CD, and join AD.

Comparing now the triangles BAC and DCA, the side AB is by supposition equal to CD, AC is

common to both, and the contained angle BAC is equal to DCA; the two triangles (I. 3.) are, therefore, equal. But this conclusion is manifestly absurd. To suppose then the inequality of AB and BC involves

B