Join the points A and H and C and H by the right lines A H and CH.

By this construction we get a A, A BH, in which the side A B is (by supposition) equal to the side D E in the ADEF; the side BH was made of the same length as the side BC or EF; and the ABH was made equal to the That is to say, the two As ABH and DEF have two sides of the one, with the between them, equal respectively to two sides of the other with the between them.

DEF.

It follows therefore that these two As are equal in every respect.

Among these respects is that the base A H is equal to the base D F.

It now remains that we should show that the line A H is less than the line A C, which is done by showing that in the AHC, the AHC is greater than the АСН.

The line BH was made of the same length as the line B C.

Consequently, in the ▲ HBC, the s BHC and BCH, which are opposite the equal sides, are themselves equal.

Now, the AHC is greater than the BHC, and is therefore greater than the BCH, which is equal to the BH C.

Moreover, the ACH is less than the BCH, and is therefore less than the AHC, which is greater than the BC H.

But it was proved in the 19th proposition that if one of a is greater than another, the side which is opposite the greater is greater than the side which is opposite the less.

Consequently, in the AA HIC, the side AC, which is opposite the larger, AHC, is greater

than the side A H, which is opposite the smaller ▲, АСН.

Now it was proved before, that the line A H is equal to the line DF.

Therefore, the line A C, which is greater than A H, is also greater than D F.*

PROPOSITION XXV.

If two triangles have two sides of the one equal to two sides of the other, each to each, but the base (or third side) of the one greater than the base (or third side) of the other, the angle contained by the two sides of the one which has the greater base will be greater than the angle contained by the corresponding two sides of the other.

The proof of this proposition is of the indirect kind. No construction is made use of for comparing the s; but it is shown that every supposition respecting the relative size of the two 8, except that stated in the enunciation of the

*Let this proposition next be gone through with the following variations in the position of the figures:

No alteration will be necessary in the wording of the proposition, while the letters are arranged as above.

proposition, must be impossible, because it would oblige us to admit something which is contradicted by what we know to be true. (See p. 28.)

For the proof of the proposition we must know,— 1. That if two As have two sides of the one equal to two sides of the other, each to each, and have likewise the included between the said sides of the one equal to the included between the said sides of the other, then those two As are also equal in every other respect, and (among those respects) have their bases equal. (Prop. IV.)

2. That if two As have two sides of the one equal to two sides of the other, each to each, but the

included between the said sides of the one greater than the included between the said sides of the other, then the base of that A which has the greater is greater than the base of that which has the smaller. (Prop. XXIV.)

Let ABC and DEF be two As in which the sides AB and BC are equal to the sides DE and EF, each to each, and the side AC greater than the side DF. It has to be shown that the ABC is greater than the DEF.

Proof. One of the three following alternatives must be true, and A only one can be true.

B

E

C D

F

A B C is less than the DEF.

Now, the A B C cannot be equal to the DEF, for, if it were, we should have the two As, ABC and DEF, with two sides and the included of the one

equal respectively to two sides and the included of the other. Consequently (according to the fourth proposition), the base A C would be equal to the base DF.

But this cannot be the case, for we know, to start with, that the base A C is greater than the base D F. Consequently, the ▲ A B C cannot be equal to the DEF.

Again, the ABC cannot be less than the DEF, for, if it were, then in the two As, A B C and DEF, we should have the two sides AB and BC equal respectively to the two sides DE and EF, but the DEF greater than the ABC. Consequently (according to the 24th proposition), the base DF would be greater than the base A C.

But this is impossible, for we know, to start with, that D F is not greater than A C.

It follows, therefore, that since the ABC cannot be equal to or less than the DEF, it must be greater than it.

PROPOSITION XXVI.

If two triangles have two angles in the one equal to two angles in the other, each to each, and have likewise one side in the one equal to a side in the other similarly placed with respect to the equal angles, then those triangles will also be equal in every other respect.

For the construction employed in this proposition we must be able,—

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

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

For the proof of the proposition we must know,-1. That magnitudes which are equal to the same, are equal to one another. (Axiom I.)

2. That if two As have two sides and the between them in the one, equal respectively to two sides and the between them in the other, those As are also equal in every other respect. (Prop. IV.)

3. That if one side of a triangle be produced, the exterior is greater than either of the two interior opposite angles. (Prop. XVI.)

This proposition divides itself into two cases :1. When the equal sides are opposite neither of the equals.

2. When the equal sides are opposite respectively to one or other of the pairs of equals.

The proof of the proposition is (in part) of the indirect kind. It is shown first that we should be compelled to admit an absurdity, if it were supposed possible that one of the other pairs of sides in the As were unequal, and then, by the application of the fourth proposition, it is proved that the two As are equal in every other respect. 1. Suppose ABC and DEF to be two As, with respect to which we know that theCAB is equal to the ZEDE; the CBA to the FED, and the side AB equal to the side DE (the sides AB and DE being opposite A

C

F

да

to neither of the pairs of equal ≤s).

B D

E

It has first to be shown that, besides the sides A B and DE, the As have also another pair of sides equal.

Take for instance, the sides A C and D F. If these sides were not equal, one would be the greater of the two. Suppose A C were greater than D F.

We should then be able to cut off from A C a part, A L, equal to D F.