Exercise.

D is a point in the base of a triangle ABC; and E is a point within the triangle ADB ; shew that the angle AEB is greater than the angle DAC.

PROPOSITION XXII. PROBLEM.

To make a triangle of which the sides shall be equal to three given straight lines, any two of which are greater than the third. Let A, B, C be three given straight lines, any two of them being greater than the third;

it is required to make a triangle whose sides shall be equal to A, B, C, each to each.

A-
B

E

F

G H

Construction.

Take a straight line DE terminated at D, but of unlimited length towards E:

in DE make DF equal to A, FG equal to B, and GH equal to C: (prop. 3)

with F as centre, and at the distance FD describe the circle DKL ; with G as centre and at the distance GH describe the circle KHL join KF, KG:

then shall the triangle KFG have its sides equal to A, B, C.

Demonstration.

Because F is the centre of the circle DKL,
therefore FK is equal to FD; (def. 11)

but FD is equal to A; (const.): therefore FK is equal to A. (ax. 1) Again, because G is the centre of the circle KHL, therefore GK is equal to GH; (def. 11)

but GH is equal to C; (const.): therefore GK is equal to C. (ax. 1) Also FG is equal to B. (const.)

Therefore a triangle KFG has been made with its sides

KF, FG, GK equal to A, B, C, each to each.

Q.E.F.

Exercises.

1. Shew that with the same figure another triangle may be made with sides equal to A, B, C.

2. Why is the condition as to the length of the sides necessary?

3. What form would the figure take if the given lines were all equal?

PROPOSITION XXIII. PROBLEM.

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

Let A be the given point in the given straight line AB, and let DCE be the given rectilineal angle;

it is required to make at A, in AB, an angle equal to DCE.

In CD, CE take any points D, E ; join DE.

Make the triangle AFG, the sides of which shall be equal to CD,

CE, DE; so that AF shall be

equal to CD, AG to CE, and

FG to DE. (prop. 22)

Then shall the angle FAG be equal to the angle DCE.

Demonstration.

Because in the triangles FAG, DCE,

AF is equal to CD, and AG to CE, and FG to DE, therefore the angle FAG is equal to the angle DCE.

(prop. 8).

Q.E.F.

Exercises:

1. Make a triangle with two of its sides equal to two given straight lines and having the angle contained by them equal to a given angle.

2. Make a triangle having two of its angles equal to two given angles, and the side adjacent to them equal to a given straight line.

3. Make a triangle having given one side and an angle adjacent to it and the sum of the other two sides.

4. If one angle of a triangle is equal to the sum of the other two, the riangle can be divided into two isosceles triangles.

PROPOSITION XXIV. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them, of the other; the base of that which has the greater angle shall be greater than the base of the other.

Let ABC, DEF be two triangles in which AB is equal to DE, and AC equal to DF,

but the angle BAC greater than the angle EDF ;
then shall the base BC be greater than the base EF.

Of the two sides DE, DF, let DE be not greater than DF. At the point D in the straight line ED make the angle EDG equal to the angle BAC; (prop. 23)

also make DG equal to DF or AC;

(prop. 3).

Join FG.

Demonstration.

In the two triangles ABC, DEG ;

AB is equal to DE, AC to DG,

and the angle BAC to the angle EDG,

therefore the base BC is equal to the base EG. (prop. 4) Again, in the triangle DFG, since DF is equal to DG, therefore the angle DFG is equal to the angle DGF; (prop. 6) therefore the angle EFG is greater than the angle DGF; (ax. 9) much more then is the angle EFG greater than the angle EGF.

(ax. 9) Now in any triangle the greater angle has the greater side opposite to it; (prop. 19)

therefore the side EG is greater than the side EF.
But it has been shewn that BC is equal to EG ;

therefore also BC is greater than EF.

Q.E.D.

Exercises.

1. Why is it necessary to state in the construction that DE is not greater than DF?

2. A, B, C are three points taken in order in the circumference of a semicircle; shew that the straight line AC is greater than the straight line AB.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle contained by the sides of that which has the greater base shall be greater than the angle contained by the sides equal to them, of the other.

Let ABC, DEF be two triangles in which AB is equal to DE, and AC equal to DF,

but BC greater than EF;

then shall the angle BAC be greater than the angle EDF.

B

D

E

F

Demonstration.

If the angle BAC is not greater than the angle EDF,
it must either be equal to it or less than it :

if it were equal to it, then BC would be equal to EF;
if it were less than it, then BC would be less than EF;

(prop. 4) (prop. 24)

but BC is greater than EF; (hyp.) therefore the angle BAC is neither equal to, nor less than the angle EDF; that is, the angle BAC is greater than the angle EDF.

Exercises.

Q.E.D.

1. D is the middle point of a straight line AB, and C a point above AB at a greater distance from A than it is from B; shew that CDA is an obtuse angle.

2. A, B, C are three points on a semicircle whose centre is 0, and AB is greater than BC; shew that the angle AOB is greater than the angle BOC.

PROPOSITION XXVI. THEOREM.

If two triangles have two angles of the one equal to two angles of the other, each to each, and one side equal to one side, namely, either the sides adjacent to the equal angles, or sides which are opposite to equal angles in each; then shall the triangles be equal in all respects.

Let ABC, DEF be two triangles in which the angle ABC is equal to the angle DEF, and the angle ACB to the angle DFE, and (in case I.) the side BC equal to the side EF ; then shall the triangles be equal in all respects.

If AB is not equal to DE, one of them must be greater than the other; suppose AB to be the greater, and from it cut off BG equal to DE: (prop. 3).

Join GC.

Demonstration.

In the triangles GBC, DEF,

GB is equal to DE, and BC to EF, and the angle GBC to the angle DEF;

therefore the triangles are equal in all respects; (prop. 4)
and therefore the angle GCB is equal to the angle DFE:
but the angle ACB is equal to the angle DFE; (hyp.)
therefore the angle GCB is equal to the angle ACB;
the less angle equal to the greater, which is impossible.
Hence AB is not unequal to DE, i.e., it is equal to it.
Now in the triangles ABC, DEF,

AB is equal to DE, and BC to EF, and the angle ABC to the

angle DEF,

therefore the triangles ABC, DEF are equal in all respects.

(prop. 4). Q.E.D.