Page images

tirely coincide with it; and consequently the points A and C resting on D and F, the straight lines AC and DF will also coincide. Wherefore, the one triangle being thus perfectly adapted to the other, a general equality must obtain between them: The third sides AC and DF are equal, and the angles BAC, BCA opposite to BC and BA are equal respectively to EDF and EFD, which the corresponding sides EF and ED subtend.


At a point in a straight line, to make an angle equal to a given angle.

At the point D in the given straight line DE to form an angle equal to the given angle BAC.

In the sides AB and AC of the given angle, assume the points G and H, join GH, from DE cut off DI equal to AG, and on DI constitute (I. 1.) a triangle DKI, having the sides DK and IK equal G to AH and GH: EDF is the B/ angle required.

For all the sides of the triangles GAH and IDK being respectively equal, the angles opposite to the equal sides must be likewise equal (I. 2.), and consequently IDK is equal to GAH.

Cor. If the segments AG, AH be taken equal, the construction will be rendered simpler and

more commodious.

Scholium. By the successive application of this problem, an angle may be continually multiplied. Two circles CEG and ADF being described from the vertex B of the given angle with

radii BC and AB equal to its sides, and the base AC

being repeated between those circumferences; a multitude of triangles are thus formed, all of them equal to the original triangle ABC. Consequently the angle ABD is double of ABC, ABE triple, ABF quadruple, ABG quintuple, &c.

If the sides AB and BC of the given angle be supposed equal, only one circle will 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 swell, and its expanding side make repeated revolutions.


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 the isosceles triangle DEF (I. 1.), 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, and consequently the angle DBF is equal to EBF.


Cor. 1. It is evident that BG, the production of BF, divides the reversed angle ABC into two equal angles DBG and EBG.-The position of BG might also be determined, by the vertex of an isosceles triangle erected above DE and with sides greater than DB or EB.

Cor. 2. Hence the mode of drawing a perpendicular from a given point B in the straight line AC; for the an

gle ABC which the opposite segments 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 construct

[blocks in formation]

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


To let fall a perpendicular upon a straight line, from a given point without it.

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

In AB take the point D, and with the distance DC describe a circle; and in the same line take another point E, and with distance EC describe a second circle intersecting the former in F; join CF, cross

ing the given line in G: CG is perpendicular to AB.

For the triangles DCE and DFE have the side DC equal to DF, CE to FE, and 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 is common, and the contained angles CDG and FDG are proved to be equal; these 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.

[ocr errors]


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.



For the sides AC and AD of the triangle CAD being respectively equal to CB 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 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.


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 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. S.) and consequently the angle BAD is equal to BCD. Cor. Every equilateral triangle is also equiangular.



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 the part 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 a contradiction; and consequently those sides must be equal.

Cor. Every equiangular triangle is also equilateral.


The exterior angle of a triangle is greater than either of the 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 D (I. 7.), draw AD, and produce it until DE be equal to AD, and join EC.

The triangles ADB and CDE have by construction the side DA equal to DE, the side DB to

DC, and the vertical angle BDA




is equal to CDE (Def. 10.); these triangles are, therefore,


« PreviousContinue »