To describe a triangle having its sides equal to three given straight lines, any two of which are together greater than the third.

Let A, B, C be the three given straight lines, of which any two are together greater than the third.

It is required to describe a triangle of which the sides shall be equal to A, B, C.

Construction. Take a straight line DE terminated at the point D, but unlimited towards E.

Make DF equal to A, FG equal to B, and GH equal to C.

I. 3.

With centre F and radius FD, describe the circle DLK. With centre G and radius GH, describe the circle MHK cutting the former circle at K.

Join FK, GK.

Then shall the triangle KFG have its sides equal to the

three straight lines A, B, C.

Proof. Because F is the centre of the circle DLK,

therefore FK is equal to FD:
but FD is equal to A;

therefore also FK is equal to A.

Again, because G is the centre of the circle MHK,

therefore GK is equal to GH:

but GH is equal to C;

therefore also GK is equal to C.

And FG is equal to B.

Def. 15.

Constr.

Ax. 1.

Def. 15.

Constr.

Ax. 1.

Constr.

Therefore the triangle KFG has its sides KF, FG, GK equal

respectively to the three given lines A, B, C.

Q.E.F.

PROPOSITION 23. PROBLEM.

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

Let AB be the given straight line, and A the given point in it, and let LCM be the given angle.

It is required to draw from A a straight line making with AB an angle equal to the given angle DCE.

Construction. In CL, CM take any points D and E;

and join DE.

From AB cut off AF equal to CD.

I. 3.

I. 22.

On AF describe the triangle FAG, having the remaining

sides AG, GF equal respectively to CE, ED.

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

Proof.

Because

For in the triangles FAG, DCE,

FA is equal to DC,

and AG is equal to CE;

Constr.

Constr.

and the base FG is equal to the base DE: Constr. therefore the angle FAG is equal to the angle DCE. I. 8. That is, AG makes with AB, at the given point A, an angle equal to the given angle DCE.

Q.E.F.

EXERCISE.

On a given base describe a triangle, whose remaining sides shall be equal to two given straight lines. Point out how the construction fails, if any one of the three given lines is greater than the sum of the other two.

PROPOSITION 24. 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 greater than the angle contained by the corresponding sides of the other; then the base of that which has the greater angle shall be greater than the base of the other.

да

CE

Let ABC, DEF be two triangles, in which
the side BA is equal to the side ED,
and the side AC is equal to the side DF,
but the angle BAC is 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 that which is not greater than the other.*

Construction. At D in the straight line ED, and on the same side of it as DF, make the angle EDG equal to the angle BAC.

Make DG equal to DF or AC;
and join EG, GF.

Proof. Then in the triangles BAC, EDG,

Because

BA is equal to ED,

and AC is equal to DG,

contained angle EDG;

I. 23.

I. 3.

Hyp. Constr.

also the contained angle BAC is equal to the

therefore the triangle BAC is equal to the triangle all respects:

so that the base BC is equal to the base EG.

Constr.

EDG in
I. 4,

Again, in the triangle FDG,
because DG is equal to DF,

therefore the angle DFG is equal to the angle DGF. 1.5. But the angle DGF is greater than its part the angle EGF; therefore also the angle DFG is greater than the angle EGF; still more then is the angle EFG greater than the angle EGF. And in the triangle EFG,

because the angle EFG is greater than the angle EGF, therefore the side EG is greater than the side EF; I. 19. but EG was shewn to be equal to BC; therefore BC is greater than EF.

Q.E.D.

*The object of this step is to make the point F fall below EG. Otherwise F might fall above, upon, or below EG; and each case would require separate treatment. But as it is not proved that this condition fulfils its object, this demonstration of Prop. 24 must be considered defective.

An alternative construction and proof are given below.

Construction. At D in ED make the angle EDG equal to the angle BAC; and make DG equal to DF. Join

EG.

Then, as before, it may be shewn that the triangle EDG=the triangle BAC in all respects. Now if EG passes through F, then EG is greater than EF; that is, BC is greater than EF.

But if not, bisect the angle FDG by DK, meeting EG at K. Join FK.

Proof. Then in the triangles FDK, GDK,

Because

FD=GD,

E

But in the triangle EKF, the two sides EK, KF are greater than EF;

that is, EK, KG are greater than EF.

Hence ÉG (or BC) is greater than EF.

PROPOSITION 25. THEOREM.

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

да

Let ABC, DEF be two triangles in which the side BA is equal to the side ED, and the side AC is equal to the side DF, but the base BC is greater than the base EF.

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

Proof. For if the angle BAC be not greater than the angle EDF, it must be either equal to, or less than the angle EDF.

But the angle BAC is not equal to the angle EDF, for then the base BC would be equal to the base EF; I. 4. but it is not.

Hyp.

Neither is the angle BAC less than the angle EDF, for then the base BC would be less than the base EF; I. 24.

but it is not.

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. Q.E.D.

EXERCISE.

In a triangle ABC, the vertex A is joined to X, the middle point of the base BC; shew that the angle AXB is obtuse or acute, according as AB is greater or less than AČ.