Join the points F and B by the right line F B.

We thus get two As BEF and CEA, in which the sides B E and CE are equal (being the equal parts into which the line BC has been divided); AE and EF are equal (EF having been cut off of the same length as A E), and the BEF is equal to the CEA, because they are vertically opposites, formed by the intersection of the lines BC and A F. So that the two As BEF and CE A have two sides and the included of the one equal respectively to two sides and the included of the other.

These two As, therefore, are also equal to each other in every other respect. (Prop. IV.)

Among these respects is that the ECA is equal EBF.

to the

But the fore, the

EBF is less than the

CBD. There-. ECA, which is equal to the EBF, is

also less than the CBD.

It has thus been shown that the exterior

CBD

is greater than that one of the interiors which is opposite to the side produced.

It still remains to be shown that the CAB is also less than the CBD.

The CAB cannot be compared directly with the CBD; but, by producing the line CB to any point G, we get an ABG equal to the CBD (being vertically opposite to it), with which the CAB may be compared in the same way as that in which the ACB was compared with the CBD.

right line CL. From the pro

For this purpose bisect the line AB in the point L. Join the points C and L by the Produce the line CL to any length. duced part cut off a part, LH, equal to the line C L. Join the points H and B by the right line H B.

We thus get two As, ALC and BLH, in which

the side AL is equal to the side BL (AB being bisected in the point L); the side CL is equal to the side LH (LH having been cut off of the same length

as CL); and the s ALC and BLH are equal, being vertically opposite. That is to say, two sides and the be-D tween them in the ▲ ALC are equal respectively to two sides and the between them in the A BLH.

A

H

G

It follows, therefore (Prop. IV.), that the As ALC and BLH are also equal in every other respect.

Among these respects is that the CAL is equal to the LBH.

But the LBH is less than the ABG (being a part of it).

Therefore the

But the

< CBD.

CAL is also less than the ABG. ABG is equal to the vertically opposite

Therefore the

is also less than the

CAL, being less than the ABG,
CBD.

It has thus been shown that the Zs ACB and CAB are each less than the exterior

by producing the side A B.

CBD, formed

Let this proposition now be demonstrated with the following variations of the figure

с

:

1. Produce the side BA to any point M, and prove that the ACB is less than the CAM. The following directions will suffice: 1. Bisect the line A C in the point P. 2. Join B and P by a straight line. 3. Produce the line BP and cut off a part, P Q, equal to PB. 4. Join the points Q and A by

M

A

B

a right line. Then show that the ▲ CPB is equal in all respects to the ▲ PQA. And so, that the

Then producing CA to N, show that the than the BAN, and therefore less than the

2. Produce the side BC to any point X, and show that the s CAB and C BA are X each of them less than the AC X.

ACB is

CBA is less
CAM.

C

The rule for proceeding is always the following:-When a side of the▲ has been produced, the that is to be compared with the exterior is that of the ▲ which is opposite the side produced. For making the necessary construction, bisect that side of the ▲ which is one of the sides containing the exterior . Join the bisecting point with the opposite vertex of the A. Produce this line through the point of bisection, and cut off a part equal to the part within the A. Lastly, connect the point so obtained, by a right line, with the vertex of the exterior ▲.

A

B

PROPOSITION XVII.

Any two angles of a triangle are less than two right angles.

For the proof of this proposition we must be able

to produce a given straight line to any length, and we must know,

1. That if one side of a ▲ be produced, the exterior
is greater than either of the two interior and
opposites. (Prop. XVI.)

2. That if one straight line meets another, the
adjacents which it forms with it are to-
gether equal to two rights. (Prop. XIII.)
3. That if the same quantity be added to each of
two unequal quantities, the sum of that quantity

A

and the larger of the two unequals is greater than the sum of that quantity and the smaller of the unequals. (Ax. IV.)

B

D

Let A B C be a A.

Take any two of its s, say CA B and CBA.

Produce one of the sides of the A, so as to form an exterior that shall be adjacent to one or other of SCAB and CBA. As, for example, produce the side A B to the point D.

the two

It follows from the last proposition that the CBD is greater than the CAB. If the CBA be added to each of these unequals, the sum of the Zs CBD and CBA will be greater than the sum of the s CAB and C B A.

But the s CBD and CBA are adjacent s, formed by the straight line CB meeting the straight line A D, and are therefore equal to two right /s.

Consequently, the Дs CAB and C B A are together less than two rights.

The proposition might have been proved equally well by either of the following constructions :

АДД

B

A

B

A

B

In each case we should have an exterior greater than one or other of the two givens of the ▲; enabling us to show that the sum of the two givens, being less than that of one of them and the exterior, is less then two rights.

Whichever two s of the ▲ are selected, it does not matter which side of the ▲ is produced, provided it is produced in a direction passing through the vertex of one of the two s fixed upon.

PROPOSITION XVIII.

If one side of a triangle be greater than another, the angle opposite the greater side is greater than the angle opposite the less.* For the construction requisite in this proposition, we must be able,

1. From the greater of two straight lines to cut off a part equal to the less. (Prop. III.)

2. To join two given points by a straight line. (Post. I.)

To prove the proposition by the aid of the construction, we must know that,—

1. If two sides of a ▲ are equal, the s opposite to those sides are also equal. (Prop. V.)

2. If one side of a ▲ be produced, the exterior ▲ is greater than either of the two interior opposite 8. (Prop. XVI.)

Let ABC be a ▲ in which the

side C B is longer than the side CA. We have to show that the

is larger than the CBA.

CAB

From the greater side, CB, cut off a part, CD, equal to the less.

D

B

* The enunciation of this proposition, as it stands in the common editions of Euclid, is almost sure to mislead the beginner. It runs :-"The greater side of every ▲ is opposite the greater." This ambiguous statement is always to be interpreted in the form given above.