(Edited by Dionysius Lardner, 1855)

Proposition XXVI. Theorem.

[Euclid, ed. Lardner, 1855, on `Google Books`]

(105) | If two triangles (B A C, D E F) have two angles of the one respectively equal to two angles of the other (B to D and C to F), and a side of the one equal to a side of the other similarly placed with respect to the equal angles, the remaining sides and angles are respectively equal to one another. |

First let the equal sides be B C and D F, which lie between the equal angles; then the side B A is equal to the side D E.

For if it be possible, let one of them B A be greater than the other; make B G equal to D E, and join C G.

In the triangles G B C, E D F the sides G B, B C are respectively equal to the sides E D, D F (const.), and the angle B is equal to the angle D (hyp.), therefore the angles B C G and D F E are equal (IV); but the angle B C A is also equal to D F E (hyp.) therefore the angle B C G is equal to B C A (51), which is absurd: neither of the sides B A and D E therefore is greater than the other, and therefore they are equal, and also B C and D F are equal (IV), and the angles B and D; therefore the side A C is equal to the side E F, as also the angle A to the angle E (IV).

Next, let the equal sides be B A and D E, which are opposite to the equal angles C and F, and the sides B C and D F, shall also be equal.

For if it be possible, let one of them B C be greater than the other; make B G equal to D F, and join A G.

In the triangles A B G, E D F, the sides A B, B G are respectively equal to the sides E D, D F (const.), and the angle B is equal to the angle D (hyp.); therefore the angles A G B and E F D are equal (IV); but the angle C is also equal to E F D, therefore A G B and C are equal, which is absurd (XVI). Neither of the sides B C and D F is therefore greater than the other, and they are consequently equal. But B A and D E are also equal, as also the angles B and D; therefore the side A C is equal to the side E F, and also the angle A to the angle E (IV).

Book I: Euclid, Book I (ed. Dionysius Lardner, 11th Edition, 1855)

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This proposition in other editions:

It is evident that the triangles themselves are equal in every respect.

^{★}★^{★}(106) Cor. 1.—From this proposition and the principles previously established, it easily follows, that a line being drawn from the vertex of a triangle to the base, if any two of the following equalities be given (except the first two), the others may be inferred.1° The equality of the sides of the triangle.

2° The equality of the angles at the base.

3° The equality of the angles under the line drawn, and the base.

4° The equality of the angles under the line drawn, and the sides.

5° The equality of the segments of the base.

Some of the cases of this investigation have already been proved (74), (75), (76). The others present no difficulty, except in the case where the fourth and fifth equalities are given to infer the others. This case may be proved as follows.

If the line A D which bisects the vertical angle (A) of a triangle also bisect the base B C, the triangle will be isosceles; for produce A D so that D E shall be equal to A D, and join E C. In the triangles D C E and A D B the angles vertically opposed at D are equal, and also the sides which contain them; therefore (IV) the angles B A D and D E C are equal, and also the sides A B and E C. But the angle B A D is equal to D A C (hyp.); and therefore D A C is equal to the angle E, therefore (VI) the sides A C and E C are equal. But A B and E C have already been proved equal, and therefore A B and A C are equal.

^{★}★^{★}(107) The twenty-sixth proposition furnishes the third criterion which has been established in the Elements for the equality of two triangles. It may be observed, that in a triangle there are six quantities which may enter into consideration, and in which two triangles may agree or differ; viz. the three sides and the three angles. We can in most cases infer the equality of two triangles in every respect, if they agree in any three of those six quantitieswhich are independent of each other. To this, however, there are certain exceptions, as will appear by the following general investigation of the question.When two triangles agree in three of the six quantities already mentioned, these three must be some of the six following combinations:

1° Two sides and the angle between them.

2° Two angles and the side between them.

3° Two sides, and the angle opposed to one of them.

4° Two angles, and the side opposed to one of them.

5° The three sides.

6° The three angles.

The first case has been established in the fourth, and the second and fourth in the twenty-sixth proposition. The fifth case has been established by the eighth, and in the sixth case the triangles are not necessarily equal. In this case, however, the three data are not independent, for it will appear by the thirty-second proposition, that any one angle of a triangle can be inferred from the other two.

The third is therefore the only case which remains to be investigated.

^{★}★^{★}(108) 3°To determine under what circumstance two triangles having two sides equal each to each, and the angles opposed to one pair of equal sides equal, shall be equal in all respects.Let the sides A B and B C be equal to D E and E F, and the angle A be equal to the angle D. If the two angles B and E be equal, it is evident that the triangles are in every respect equal by (IV), and that C and F are equal. But if B and E be not equal, let one B be greater than the other E; and from B let a line B G be drawn, making the angle A B G equal to the angle E. In the triangles A B G and D E F, the angles A and A B G are equal respectively to D and E, and the side A B is equal to D E, therefore (XXVI) the triangles are in every respect equal; and the side B G is equal to E F, and the angle B G A equal to the angle F. But since E F is equal to B C, B G is equal to B C, and therefore (V) B G C is equal to B C G, and therefore C and B G A or F are supplemental.(109) Hence,

if two triangles have two sides in the one respectively equal to two sides in the other, and the angles opposed to one pair of equal sides equal, the angles opposed to the other equal sides will be either equal or supplemental.^{★}★^{★}(110) Hence it follows, that if two triangles have two sides respectively equal each to each, and the angles opposed to one pair of equal sides equal, the remaining angles will be equal, and therefore the triangles will be in every respect equal, if there be any circumstance from which it may be inferred that the angles opposed to the other pair of equal sides are of the same species.(Angles are said to be of the same species when they are both acute, both obtuse, or both right).

For in this case, if they be not right they cannot be supplemental, and must therefore be equal (109), in which case the triangles will be in every respect equal, by (XXVI).

If they be both right, the triangles will be equal by (108); because in that case G and C being right angles, B G must coincide with B C, and the triangle B G A with B C A; but the triangle B G A is equal to E F D, therefore &c.

^{★}★^{★}(111) There are several circumstances which may determine the angles opposed to the other pair of equal sides to be of the same species, and therefore which will determine the equality of the triangles; amongst which are the following:If one of the two angles opposed to the other pair of equal side be right; for a right angle is its own supplement.

If the angles which are given equal be obtuse or right; for then the other angles must be all acute (91), and therefore of the same species.

If the angles which are included by the equal sides be both right or obtuse; for then the remaining angles must be both acute.

If the equal sides opposed to angles which are not given equal be less than the other sides, these angles must be both acute (XVIII).

In all these cases it may be inferred, that the triangles are in every respect equal.

It will appear by prop. 38, that if two triangles have two sides respectively equal, and the included angles supplemental, their areas are equal.

(The

areaof a figure is the quantity of surface within its perimeter).(112) If several right lines be drawn from a point to a given right line.

1° The shortest is that which is perpendicular to it.

2° Those equally inclined to the perpendicular are equal, and

vice versa.3° Those which meet the right line at equal distances from the perpendicular are equal, and

vice versa.4° Those which make greater angles with the perpendicular are greater, and

vice versa.5° Those which meet the line at greater distances from the perpendicular are greater, and

vice versa.6° More than two equal right lines cannot be drawn from the same point to the same right line.

The student will find no difficulty in establishing these principles.

^{★}★^{★}(113) If any number of isosceles triangles be constructed upon the same base, their vertices will be all placed upon the right line, which is perpendicular to the base, and passes through its middle point. This is a very obvious and simple example of a species of theorem which frequently occurs in geometrical investigations. This perpendicular is said to be thelocusof the vertex of isosceles triangles standing on the same base.