ral triangle DEF (23); then join AF, and the angle BAC is bisected by the line AF.

For, since D and E are points in the circumference of the same circle whose centre is A, AD = AE; and since DEF is an equilateral triangle, DF = EF. Therefore in the two triangles ADF, AEF, the three sides AD, DF, AF are equal to the three sides AE, EF, AF, each to each, in order; so that (24, COR.) the two triangles are equal in all respects, and the angle DAF between AD, AF, is equal to the angle EAF between AE, AF. Therefore the angle BAC is divided into two equal angles by the straight line AF.

26. PROP. IV. The angles at the base* of an isosceles triangle are equal to one another.

Let ABC be an isosceles triangle, in which the side AB= the side AC, and BC is the third side, or base; the angle ABC shall be equal to the angle ACB.

Bisect the angle BAC by the straight line AD (25), D being the point where AD meets the base BC. Then, since AB AC, and ▲ BAD = ▲ CAD, we have two triangles ABD, ACD, in which the two sides BA, AD are equal to the

D

C

two sides CA, AD, each to each, and the angle formed by the two sides of the one equal to the angle formed by the two sides of the other, .. the triangles are equal in all respects (24), and the angles are equal, each to each, to which the equal sides are opposite; and

or,

.. 4 ABD= 2 ACD,

which is the same thing, < ABC = 4 ACB.

COR. Hence every equilateral triangle is also equiangular; and, conversely, every equiangular triangle is also equilateral.

27. PROP. V. To bisect a given finite straight line, that is, to divide it into two equal straight lines.

*The base in an isosceles triangle is restricted to one side, viz. the unequal side, on which the two equal sides may be supposed to stand, except when the triangle is also equilateral, in which case any side may be taken as the base.

A given finite straight line' means a straight line fixed both in position and magnitude.

Let AB be the given straight line. Upon AB describe the equilateral triangle ABC (23); bisect the angle ACB by the straight line CD meeting AB in D (25); then AB is bisected in the point D.

For AC, CD are equal to BC, CD, each to each, and ▲ ACD = L BCD, .. the triangles ACD, BCD are equal in all respects (24); and .. AD=BD,

A

being the sides opposite to the equal angles ACD, BCD; that is, AB is divided into two equal parts in the point D.

28. PROP. VI. To draw a straight line at right angles to a given straight line* from a given point in it. Let AB be the given straight line, and C a given point in it. It is required to draw a straight line from C at right angles to AB. In AC take any point D, and with centre C, and

radius CD, describe

an arc of a circle A

D

E

B

cutting the line AB in D and E; upon DE describe the equilateral triangle DEF (23); and join FC; CF shall be at right angles to AB.

=

For, since CD CE (16), and DF = EF (23), by construction, in the two triangles DCF, ECF, DC, CF are equal to EC, CF, each to each, and the third side DF is equal to EF, .. the triangles are equal in all respects, (24, Cor.) and DCF = ¿ECF, to which the equal sides DF, EF are respectively opposite, and .. each of them is a right angle (10), that is, CF is at right angles to AB.

29. PROP. VII. To draw a straight line † perpendi

*This straight line is required to be given in position only.

† Whether a certain straight line is drawn at right angles, or perpendicular, to another straight line, depends upon the simple fact, whether it be drawn, from a known point in the latter line itself, away from the line, or from a known point without it towards the line.

cular to a given straight line of unlimited length from a given point without it.

Let AB be the given straight line, and C a given point without it, from which it is required to draw a perpendicular to AB. Take a point D on the other side of AB, and with centre C and radius CD, describe a circle cutting AB, or AB produced, in E and

G

D

F; join CE, CF, and bisect ECF by the line CG, meeting AB in G. Then CG shall be perpendicular to AB.

For CE=CF (16), and ECG L = FCG, by construction, in the triangles ECG, FCG, EC, CG are equal to FC, CG, each to each, and ECG = < FCG, ..the triangles are equal in all respects (24), and the angles equal to which equal sides are opposite, viz.

LEGC= 4 FGC,

and.. each of them is a right angle, or CG is perpendicular to AB (10).

30. PROP. VIII. The two angles which one straight line makes with another upon the one side of it are always equal to two right angles.

Let the straight line CD meet the straight line AB in the point C, and make with AB on the one side of it the angles ACD, BCD; these are always equal to two right angles.

For, if 4 ACD = ▲ BCD, then each of them is a right angle (10), and .. the two together make two right angles.

Or, if the angles ACD, BCD are unequal, from the point Cin AB draw CE at right angles to AB (28); then, supposing ACD to be the greater of the two angles ACD, BCD, it is evident that ACD is as much greater than a right angle as BCD is less, and .. that the two together are equal to two right angles.

L

COR. 1. If the two angles ACD, BCD, with the same vertex C', are together equal to two right angles, AC, and CB, are in one and the same straight line.

COR. 2. Hence, also, whatever be the number of angles in one plane, separate and distinct, on one side of AB, with a common vertex C, the sum of all these angles is equal to two right angles; and similarly on the other side of AB. So that all the angles in one plane exactly occupying the whole space round any given point are together equal to four right angles.

31. PROP. IX. If one straight line intersects* another straight line, the vertical, or opposite, angles shall be equal to one another.

Let the straight line AB intersect the straight line CD in the point E; then

LAEC=LBED,

and AED = ‹ BEC.

For, since CE meets AB, and makes with it the angles AEC, BEC, these are together equal to two right angles (30); and since BE meets CD, and

A

T

D

makes with it the angles BEC, BED, these also are equal to two right angles; .. the angles AEC, BEC together are equal to the angles BEC, and BED together; and, taking the same angle BEC from these equals, the remainders must be equal, that is,

Similarly, AEC, AED are equal to two right angles; and so also are 28 AEC, BEC;

.. 4 AED = L BEC.

COR. Hence, if the lines forming any angle be 'produced', or extended, through the vertex in the opposite direction, the new angle thus formed will be equal to the other.

32. PROP. X. If a side of a triangle be 'produced'+,

* One line intersects another when it not only meets that other, but crosses it and is continued on the other side.

+ That is, be extended, or continued indefinitely in the same straight line.

the exterior angle, thus formed by the adjacent side and the side produced, is greater than either of the interior opposite angles.

F

Let the side BC of the triangle ABC be 'produced' to any point D; the exterior angle ACD shall be greater than either of the interior opposite angles ABC, BAC.

Bisect the side AC in the point E (27), join BE, produce it to F,

making EF equal to EB B

(by describing a circle with centre E and radius EB to cut BE produced in F), and join CF.

Then in the triangles AEB, CEF, AE, EB are equal to CE, EF, each to each, by construction, and 2 AEB= = CEF (31), because they are opposite vertical angles,.. the triangles AEB, CEF are equal in all respects (24), and the angles are equal to which the equal sides are opposite, so that BAE= ECF; but ▲ ACD is greater than ▲ ECF (8), .. LACD is greater than BAE, or BAC, which is the same thing.

Similarly it may be shewn, by bisecting BC in G, joining AG, and proceeding as before, that ACD is greater than ABC.

2

33. PROP. XI. The greater side of every triangle is opposite to the greater angle.

Let ABC be a triangle, of which the side AC is greater than the side AB;

ABC shall be greater

than BCA.

With centre A and radius AB describe an arc of a circle cutting AC in

the point D; and join BD,

which will necessarily fall within the triangle ABC.

C

Then, since AB = AD, L ABD = 4 ADB (26); but LADB, the exterior angle of the triangle DBC, is greater than the interior opposite BCD (32); .. ‹ ABD is greater than BCD, or BCA, and .. à fortiori ABC is greater than 4 BCA.