PROP. XXIII. PROB.

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

Let AB be the given straight line,

and A the given point in it, and DCE the given rectilineal angle;

AA

it is required to make an angle at the given point A D

in the given straight line AB,

B

that shall be equal to the given rectilineal angle DCE.

G

the sides of which shall be equal to the three straight lines

CD, DE, EC,

so that CD be equal to AF,

CE to AG,

and DE to FG;

BOOK I.

xxii. 1.

And because DC, CE are equal to FA, AG, each to each,
and the base DE to the base FG;

the angle DCE is equal to the angle FAG.
Therefore, at the given point A in the given straight line AB,
the angle FAG is made equal to the given rectilineal angle
DCE.

Which was to be done.

[The best mode of drawing the angle required is by making CD and CE equal to each other. Then in AB,

b viii. 1.

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,

which have the two sides,

AB, AC equal to the two DE, DF,

each to each, viz. AB equal to DE, and AC to DF ;

but the angle BAC greater than the angle EDF;

the base BC is also greater than the base EF.

Of the two sides DE, DF,

let DE be the side which is not greater than the other,
and at the point D, in the straight line DE,

* xxiii. 1. makea the angle EDG equal to the angle BAC;
and make DG equal to AC or DF,

b iii. 1.

and join EG, GF.

b

Because AB is equal to DE,

and AC to DG,

the two sides BA, AC are equal to the two ED, DG,

each to each, and the angle BAC is equal to the angle EDG ; therefore the base BC is equal to the base EG;

с

DFG is greater than EGF;

and much more is the angle EFG

greater than the angle EGF;

and because the angle EFG of the triangle EFG

is greater than its angle EGF,

e

and that the greater side is opposite to the greater angle ; the side EG is therefore greater than the side EF;

but EG is equal to BC;

and therefore also BC is greater than EF.

Therefore, if two triangles, &c. Q. E. D.

e xix. 1.

PROP. XXV. THEOR.

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 also 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,

which have the two sides,

AB, AC equal to the two sides DE, DF,

each to each, viz. AB equal to DE, and AC to DF;

but the base CB greater than the base EF;

the angle BAC is likewise greater than the angle EDF. For, if it be not greater,

it must either be equal to it, or less;

BOOK I. but the angle BAC is not equal to the angle EDF, because then the base

b

E

b xxiv. 1. because then the base BC would be less than the base EF;

but it is not;

therefore the angle BAC is not less than the angle EDF; and it was shown that it is not equal to it;

therefore the angle BAC is greater than the angle EDF. Wherefore, if two triangles, &c. Q. E.D.

[It will be worth the student's while to cut out two triangles in cardboard, so as to exemplify the two last propositions.]

PROP. XXVI. THEOR.

If two triangles have two angles of one equal to two angles of the other, each to each; and one side equal to one side, viz. either the sides adjacent to the equal angles, or the sides opposite to equal angles in each; then shall the other sides be equal, each to each; and also the third angle of the one to the third angle of the other.

Let ABC, DEF be two triangles, which have the angles ABC, BCA

equal to the angles DEF, EFD,

viz. ABC to DEF, and BCA to EFD;

also one side equal to one side;

CASE I. And first let those sides be equal which are adjacent to the angles that are equal in the two triangles ; viz. BC to EF;

Therefore, because BG is equal to DE, and BC to EF,

the two sides GB, BC are equal to the two DE, EF, each to each; and the angle GBC is equal to DEF; therefore the base GC is equal to the base DF,

а

and the triangle GBC to the triangle DEF,
and the other angles to the other angles,

each to each, to which the equal sides are opposite;
therefore the angle GCB is equal to the angle DFE;
but DFE is, by the hypothesis, equal to the angle BCA,
wherefore also the angle BCG is equal to the angle BCA,
the less to the greater,

which is impossible;

therefore AB is not unequal to DE,

that is, it is equal to it;

and BC is equal to EF;

therefore the two AB, BC are equal to the two, DE, EF,

each to each; and the angle ABC is equal to DEF;

a

the base therefore AC is equal to the base DF,

and the third angle BAC to the third angle EDF.

a iv. 1.