Let ABC, DEF be two triangles.in which

the side AB is equal to the side DE,

and the side AC equal to the side DF, and also the base BC equal to the base EF;

then shall the angle BAC be equal to the angle EDF.

Let the triangle ABC be applied to the triangle DEF,
so that the point B may be on the point E,
and the straight line BC on the straight line EF;
then since BC is equal to EF, (hyp.)

therefore the point C will coincide with the point F.
And BC coinciding with EF,

BA and AC will coincide with ED and DF ;

for if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, they wil have a different situation as EG, FG: but then on the same base and on the same side of it there will be two triangles having their sides which are terminated in the one extremity of the base equal to one another, and likewise those which are terminated in the other extremity; but this is impossible: (prop. 7)

therefore the sides BA, CA cannot have any situation
such as EG, FG,

that is, if the base BC coincides with the base EF,

the sides BA, AC cannot but coincide with the sides ED, DF; and therefore the angle BAC coincides with the angle EDF,

and is equal to it. (ax. 8)

Exercises.

Q.E.D.

1. In an isosceles triangle, the straight line which joins the vertical angle to the middle point of the base, bisects the vertical angle, and is at right angles to the base.

2. If a straight line be drawn from the centre of a circle to bisect another straight line which is drawn with its extremities on the circumference of the circle, the two straight lines will be at right angles.

3. In the figure of Proposition I., if F be the other point in which the circles intersect each other, shew that the angle CAF is equal to the angle CBF.

4. If in an equilateral triangle ABC a point G can be found such that GA, GB, GC are all equal; shew that the three angles ADB, BDC, CDA are also equal, and that the angles of the triangle are bisected.

5. Prove the following construction for bisecting an angle ABC: make AB and CB of equal length; then with centre A and distance AB describe a circle, and with centre C and distance CB describe another circle, and let them intersect in B and D; then shall DB bisect the angle ABC.

6. Prove the following construction for bisecting a straight line AB: with centres A and B describe two equal circles with distances greater than half of AB, and let them cut each other in C, and D; then CD shall bisect AB at right angles.

To bisect a given rectilineal angle, that is, to divide it into two

Construction.

In AB take any point D;

from AC cut off AE equal to AD; (prop. 3)

join DE;

upon DE, on the side remote from A, describe the equilateral triangle DEF; (prop. 1) join AF.

Then shall AF bisect the angle BAC.

Demonstration.

In the triangles DAF, EAF,

since the side AD is equal to the side AE, and the side AF is common to the triangles, and the base DF is equal to the base EF;

therefore the angle DAF is equal to the angle EAF; (prop. 8) that is, the angle BAC is bisected by the straight line AF.

Q. E. F.

Exercises.

1. Divide a given angle into four equal angles.

2. Upon a given base make an isosceles triangle having each of the angles at the base half of those of an equilateral triangle.

3. ABC is a triangle in which the angle B is double of the angle C; find a point D in AC such that DB is equal to DC.

4. Find a point within an equilateral triangle which shall be equidistant from its angular points.

PROPOSITION X. PROBLEM.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let AB be the given straight line;

it is required to bisect it.

Upon AB describe an equilateral triangle ABC; (prop. 1) bisect the angle ACB by the straight line CD.

Then shall AB be bisected at the point D.

Demonstration.

(prop. 9)

Because, in the triangles ACD, BCD, AC is equal to BC, (def. 15) and CD is common to the triangles,

(constr.)
(prop. 4)

and the angle ACD is equal to the angle BCD;
therefore the base AD is equal to the base BD :
that is, the straight line AB is bisected at the point D.

Exercises.

Q.E.F.

1. If E be any point in CD, then EAB is an isosceles triangle.

2. Divide a straight line into four equal parts.

3. Find three points in the sides of an equilateral triangle, such that by being joined they shall form another equilateral triangle.

4. In the figure of Proposition I., if AB be produced both ways to meet the circles in L and M, shew that CLM is an isosceles triangle,

PROPOSITION XI. PROBLEM.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be the given straight line, and C the given point in it; it is required to draw a straight line at right angles to AB from the

from CB cut off CE equal to CD; (prop. 3)
on DE describe the equilateral triangle DEF; (prop. 1)
join CF. (post. 1)

Then shall CF, drawn from C, be at right angles to AB.

Demonstration.

Because, in the triangles DCF, ECF,
DC is equal to EC,

and CF is common to the triangles,

and the base DF is equal to the base EF,

therefore the angle DCF is equal to the angle ECF; (prop. 8) and they are adjacent angles :

but "when a straight line standing on another straight line makes the adjacent angles equal to one another, each of these angles is called a right angle;" (def. 9)

therefore each of the angles DCF, ECF is a right angle : that is, from the point C, CF has been drawn at right angles to AB.

Q.E.F.

COROLLARY.-By the help of this problem, it may be shewn that two straight lines cannot have a common segment.

For if it be possible let the straight lines GHK, GHL have a segment GH common to them.

From H draw HM at right angles to GH.

Then because KHG is a straight line,

therefore the angle KHM is equal to the angle GHM; (def. 9) and because LHG is a straight line,

therefore the angle LHM is equal to the angle GHM: (def. 9) hence the angle KHM is equal to the angle LHM; (ax. 1) the less equal to the greater, which is impossible. Therefore two straight lines cannot have a common segment.

Exercises.

1. Draw a straight line any point of which shall be equidistant from the extremities of a given straight line.

2. In a given straight line determine a point which shall be equidistant from two given points, one in the given line and the other above it.

3. Two points lie on opposite sides of a given straight line; find a point in the line equidistant from the two given points.

4. Shew by figures that the given points in questions (2) and (3) may be so situated that it will be impossible to find a point in the given line equidistant from them.

PROPOSITION XII. PROBLEM.

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

Let AB be the given straight line of unlimited length,
and let C be the given point without it;

it is required to draw a straight line perpendicular to AB
from the point C.