STRAIGHT LINES, ANGLES, AND TRIANGLES.

DEFINITION.

An equilateral triangle is one which has all its sides equal.

PROPOSITION I.

To describe an equilateral triangle upon a given straight line.

Let AB be the given straight line.

It is required to describe an equilateral triangle upon AB.

With centre A, and radius AB, describe the circle BCD; and with centre B, and radius BA, describe the circle ACE.

From the point C, in which these circles cut one another, draw the straight lines CA, CB.

Then ABC shall be an equilateral triangle,

Because A is the centre of the circle BCD, therefore AC is equal to AB;

and because B is the centre of the circle ACE, therefore BC is equal to AB.

Thus AC and BC are each of them equal to AB. Therefore AC and BC are equal to one another. Therefore AC, BC, and AB are equal to one another. Therefore the triangle ABC is equilateral.

Wherefore an equilateral triangle has been described upon AB.

WHICH WAS TO BE DONE.

PROPOSITION II.

To describe a triangle having its sides equal to three given straight lines.

Let AB, CD, and EF be the three given straight lines. It is required to describe a triangle having its sides equal to AB, CD, and EF respectively.

With centre A, and radius equal to CD, describe the circle MP;

and with centre B, and radius equal to EF, describe the circle QR.

If these circles intersect in S, join AS and BS.
Then shall ABS be the triangle required.

For AS is a radius of the circle MP, and is therefore equal to CD;

also BS is a radius of the circle QR, and is therefore equal to EF.

Wherefore a triangle ABS has been described having its sides AB, AS, BS respectively equal to the given straight lines AB, CD, EF.

QUOD ERAT FACIENDUM.

Note. If the circles do not intersect, the problem cannot be done.

PROPOSITION III.

To describe a triangle having two of its sides equal to two given straight lines, and the angle contained by those sides equal to a given angle.

A

Let A and B be the given straight lines, and PQR the given angle.

It is required to describe a triangle having two of its sides equal to A and B, and the angle contained by those sides equal to the angle PQR.

From QR cut off QS, equal to A ; and from QP cut off QT, equal to B.

Join ST.

Then shall the triangle QST be the one required.

For the triangle QST has two of its sides, QS, QT, equal to the two given straight lines A and B, and they contain the angle TQS, which is the same as the given angle PQR.

Wherefore a triangle has been described having two of its sides &c.

Q.E.F.

PROPOSITION IV.

If two sides and the included angle of one triangle be respectively equal to two sides and the included angle of another triangle, the triangles shall be equal in all respects.

In the triangles ABC, DEF,

let the two sides AB, AC, and the included angle BAC, be respectively equal to the two sides DE, DF, and the included angle EDF

Then shall the triangles ABC, DEF be equal in all respects.

For if the triangle ABC were applied to the triangle DEF so that the point A fell on the point D, and the straight line AB along DE, then would AC fall along DF, because the angle BAC is equal to the angle EDF;

also B would fall on E, because AB is equal to DE; and C on F, because AC is equal to DF.

Therefore the straight line BC would coincide with EF; for if it fell otherwise, as EKF, then two straight lines would enclose a space, which is impossible.

Therefore BC would coincide with EF, and is therefore equal to it; and the triangle ABC would coincide. with the triangle DEF, and is equal to it.

So the angle ABC would coincide with the angle DEF, and is equal to it; and the angle ACB would coincide with the angle DFE, and is equal to it.

Thus the triangle ABC is equal to the triangle DEF in all respects.

Wherefore if two sides &c.,

WHICH WAS TO BE PROVED.

DEFINITION.

An isosceles triangle is one which has two of its sides

equal.

PROPOSITION V.

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

B

Let ABC be an isosceles triangle, having the side AB equal to the side AC.

Then shall the angle ABC be equal to the angle ACB.

For if the triangle were taken up, reversed, and then placed so that A fell on its former position,

and AB along the former position of AC;

then would AC fall along that of AB;

also B would fall on that of C, because AB is equal to AC, and C on that of B for the same reason.

Therefore BC would fall on its former position.

Thus AB, BC would fall on the former positions of AC, CB;

Therefore the angle ABC is equal to the angle ACB. Wherefore the angles at the base &c.

QUOD ERAT DEMONSTRANDUM.