Let the two triangles ABC, DEF, have the angle A equal to the angle D, the angle B equal to the angle E, and the side AB equal to the side DE; then these two triangles will be identical.

A

B D E

For, conceive the triangle ABC to be placed on the triangle DEF, in such manner that the side AB may fall exactly on the equal side DE. Then, since the angle A is equal to the angle D (by hyp.), the side AC must fall on the side DF; and, in like manner, because the angle B is equal to the angle E, the side BC must fall on the side EF. Thus the three sides of the triangle ABC will be exactly placed on the three sides of the triangle DEF; consequently the two triangles are identical (ax. 9), having the other two sides AC, BC, equal to the two DF, EF, and the remaining angle C equal to the remaining angle F.

Q. E. D.

THEOREM HI

In an isosceles triangle, the angles at the base are equal. Or, if a triangle have two sides equal, their opposite angles will also be equal.

If the triangle ABC have the side AC equal to the side BC: then will the angle B be equal to the angle A.

For, conceive the angle C to be bisected, or divided into two equal parts, by the line CD, making the angle ACD equal to the angle BCD.

A D B

Then, the two triangles ACD, BCD, have two sides and the contained angle of the one, equal to two sides and the contained angle of the other, viz. the side AC equal to BC, the angle ACD equal to BCD, and the side CD common; therefore these two triangles are identical, or equal in all respects (th. 1); and consequently the angle A equal to the angle B. Q. E. D.

Corol. 1. Hence the line which bisects the vertical angle of an isosceles triangle, bisects the base, and is also perpendicular to it.

Corol. 2. Hence too it appears, that every equilateral triangle, is also equiangular, or has all its angles equal.

THEOREM IV.

When a triangle has two of its angles equal, the sides opposite to them are also equal.

If the triangle ABC, have the angle A equal to the angle B, it will also have the side AC equal to the side BC.

For, conceive the side AB to be bisected in the point D, making AD equal to DB; and join DC, dividing the whole triangle into the two triangles ACD, BCD. Also conceive the triangle ACD to be turned over upon the triangle BCD, so that AD may fall on BD. (th. 3, Cor. 1)

C

A D

Then, because the line AD is equal to the line DB (by hyp.), the point A coincides with the point B, and the point D with the point D. Also, because the angle A is equal to the angle B (by hyp.), the line AC will fall on the line EC, and the extremity C of the side AC will coincide with the extremity C of

the side BC, because DC is common to both; consequently the side AC is

equal to BC.

Q. E. D.

Corol. Hence every equiangular triangle is also equilateral.

THEOREM V.

When two triangles have all the three sides in the one, equal to all the three sides in the other, the triangles are identical, or have also their three angles equal, each to each.

Let the two triangles ABC, ABD, have their three sides respectively equal, viz. the side AB equal to AB, AC to AD, and BC to BD; then shall the two triangles be identical, or have their angles equal, viz. those angles that are opposite to the equal sides; namely, the angle BAC to the angle BAD, the angle ABC to the angle ABD, and the angle C to the angle D.

For, conceive the two triangles to be joined together by their longest equal sides, and draw the line CD.

D

B

Then, in the triangle ACD, because the side AC is equal to AD (by hyp.), the angle ACD is equal to the angle ADC (th. 3). In like manner, in the triangle BCD, the angle BCD is equal to the angle BDC, because the side BC is equal to BD. Hence then, the angle ACD being equal to the angle ADC, and the angle BCD to the angle BDC, by equal additions the sum of the two angles, ACD, BCD, is equal to the sum of the two ADC, BDC, (ax. 2), that is, the whole angle ACB equal to the whole angle ADB.

Since, then, the two sides AC, CB, are equal to the two sides AD, DB, each to each, (by hyp.), and their contained angles ACB, ADB, also equal, the two triangles ABC, ABD, are identical (th, 1), and have the other angles equal, viz. the angle BAC to the angle BAD, and the angle ABC to the angle ABD. Q. E. D.

THEOREM VI.

When one line meets another, the angles which it makes on the same side of the other, are together equal to two right angles.

Let the line AB meet the line CD: then will the two angles ABC, ABD, taken together, be equal to two right angles.

E

B

D

For, first, when the two angles ABC, ABD, are equal to each other, they are both of them right angles (def. 15). But when the angles are unequal, suppose BE drawn perpendicular to CD. Then, since the two angles EBC, EBD, are right angles (def. 15), and the angle EBD is equal to the two angles EBA, ABD, together (ax. 8), the three angles, EBC, EBA, and ABD, are equal to two right angles.

But the two angles EBC, EBA, are together equal to the angle ABC (ax. 8). Consequently, the two angles ABC, ABD, are also equal to two right angles. Q. E. D.

Corol. 1. Hence also, conversely, if the two angles ABC, ABD, on both

sides of the line AB, make up together two right angles, then CB and BI) form one continued right line CD.

Corol. 2. Hence, all the angles which can be made, at any point B, by any number of lines, on the same side of the right line CD, are, when taken all together, equal to two right angles.

Corol. 3. And, as all the angles that can be made on the other side of the line CD are also equal to two right angles; therefore, all the angles that can be made quite round a point B, by any number of lines, are equal to four right angles.

Corol. 4. Hence, also, the whole circumference of a circle, being the sum of the measures of all the angles that can be made about the centre F (def. 57), is the measure of four right angles. Consequently, a semicircle, or 180 degrees, is the measure of two right angles; and a quadrant, or 90 degrees, the measure of one right angle.

D

THEOREM VII.

When two lines intersect each other, the opposite angles are equal. Let the two lines AB, CD, intersect in the point E; then will the angle AEC be equal to the angle BED, and the angle AED be equal to the angle ĈEB.

For, since the line CE meets the line AB, the two angles AEC, BEC, taken together, are equal to two right angles (th. 6).

In like manner, the line BE, meeting the line CD, makes the two angles BEC, BED, equal to two right angles.

B

E

D

Therefore, the sum of the two angles AEC, BEC, is equal to the sum of the two BEC, BED (ax. 1).

And if the angle BEC, which is common, be taken away from both these, the remaining angle AEC will be equal to the remaining angle BED (ax. 3). And in like manner it may be shown, that the angle AED is equal to the opposite angle BEC.

THEOREM VIII.

When one side of a triangle is produced, the outward angle is greater than either of the two inward opposite angles.

Let ABC be a triangle, having the side AB produced to D; then will the outward angle CBD be greater than either of the inward opposite angles A or C.

For, conceive the side BC to be bisected in the point E, and draw the line AE, producing it till EF be equal to AE; and join BF.

=

A

B

D

Then, since the two triangles AEC, BEF, have the side AE the side EF, and the side CE = the side BE (by suppos.), and the included or opposite angles at E also equal (th. 7), therefore, those two triangles are equal in all respects (th. 1), and have the angle C = the corresponding angle EBF. But the angle CBD is greater than the angle EBF; conse quently, the said outward angle CBD is also greater than the angle C.

In like manner, if CB be produced to G, and AB be bisected, it may be shown that the outward angle ABG, or its equal CBD, is greater than the other angle A.

THEOREM IX.

The greater side, of every triangle, is opposite to the greater angle; and the greater angle opposite to the greater side.

Let ABC be a triangle, having the side AB greater than the side AC; then will the angle ACB, opposite the greater side AB, be greater than the angle B, opposite the less side AC.

A

D B

For, on the greater side AB, take the part AD equal to the less side AC, and join CD. Then, since BCD is a triangle, the outward angle ADC is greater than the inward opposite angle B (th. 8). But the angle ACD is equal to the said outward angle ADC, because AD is equal to AC (th. 3). Consequently, the angle ACD also is greater than the angle B. And since the angle ACD is only a part of ACB, much more must the whole angle ACB be greater than the angle B. Q. E. D.

Again, conversely, if the angle C be greater than the angle B, then will the side AB, opposite the former, be greater than the side AC, opposite the latter. For, if AB be not greater than AC, it must be either equal to it, or less than it. But it cannot be equal, for then the angle C would be equal to the angle B (th. 3), which it is not, by the supposition. Neither can it be less, for then the angle would be less than the angle B, by the former part of this; which is also contrary to the supposition. The side AB, then, being neither equal to AC, nor less than it, must necessarily be greater. Q. E. D.

THEOREM X.

The sum of any two sides of a triangle is greater than the third side.

Let ABC be a triangle; then will the sum of any two

of its sides be greater than the third side, as for in

stance, ACCB greater than AB.

But the angle ABD consequently, it must

A

B

For, produce AC till CD be equal to CB, or AD equal to the sum of the two AC+ CB; and join BD:—Then, because CD is equal to CB (by constr.), the angle D is equal to the angle CBD (th. 3). is greater than the angle CBD, also be greater than the angle D. And, since the greater side of any triangle is opposite to the greater angle (th. 9), the side AD (of the triangle ABD) is greater than the side AB. But AD is equal to AC and CD, or AC and CB, taken together (by constr.); therefore, AC+ CB is also greater than AB Q E. D.

Corol. The shortest distance between two points, is a single right line drawn from the one point to the other.

THEOREM XI.

The difference of any two sides of a triangle, is less than the third side.

Let ABC be a triangle; then will the difference of any two sides, as AB AC, be less than the third side BC.

For, produce the less side AC to D, till AD be equal to the greater side AB, so that CD may be the difference of the two sides AB AC; and join BD. Then, because AD is equal to AB (by constr.), the opposite angles D and ABD are equal (th. 3). But the angle

D

B

CBD is less than the angle ABD, and consequently also less than the equal angle D. And since the greater side of any triangle is opposite to the greater angie (th. 9), the side CD (of the triangle BCD) is less than the side BC. Q. E. D. Otherwise. Set off upon AB a distance AI equal to

AC.

Then (th. 20) AC + CB is greater than AB, that is, greater than AI + IB. From these, take away the equal parts, AC, AI, respectively; and there remains CB greater than IB. Consequently, IB is less than CB. Q. E. D.

C

THEOREM XII

When a line intersects two parallel lines, it makes the alternate angles equal to each other.

Let the line EF cut the two parallel lines AB, CD; then will the angle AEF be equal to the alternate angle EFD.

For if they are not equal, one of them must be greater than the other; let it be EFD for instance which is the greater, if possible; and conceive the lina FB to be drawn, cutting off the part or angle EFB equal to the angle AEF, and meeting the line AB in the point B.

A

E

B

C

F

D

Then, since the outward angle AEF, of the triangle BEF, is greater than the inward opposite angle EFB (th. 8); and since these two angles also are equal (by the constr.) it follows, that those angles are both equal and unequal at the same time: which is impossible. Therefore the angle EFD is not unequal to the alternate angle AEF, that is, they are equal to each other. Q E. D.

Corol. Right lines which are perpendicular to one, of two parallel lines, are also perpendicular to the other.

THEOREM XIII.

When a line, cutting two other lines, makes the alternate angles equal to each other, those two lines are parallel.

Let the line EF, cutting the two lines AB, CD, make the alternate angles AEF, DFE, equal to each other; then will AB be parallel to CD.

C

E

For if they be not parallel, let some other line, as FG, be parallel to AB. Then, because of these parallels, the angle AEF is equal to the alternate angle EFG (th. 1z). But the angle AEF is equal to the angle EFD (by hyp.) Therefore the angle EF1) is equal to the angle EFG (ax. 1); that is, a part is equal to the whole, which is impossible. Therefore no line but CD can be parallel to AB. QE D.