PROPOSITION I. PROBLEM.

To describe an equilateral triangle upon a given finite straight line.

Let AB be the given straight line; it is required to describe an equilateral triangle upon it.

From the centre A, at the distance AB, describe (3. Postulate) the circle BCD, and from the centre B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines (1. Post.)CA, CB to the points A, B; ABC is an equilateral triangle.

D

A

C

Because the point A is the centre of the circle BCD, AC is equal (11. Definition) to AB; and because the point B is the centre of the circle ACE, BC is equal to AB: But it has been proved that CA is equal to AB; therefore CA, CB are each of them equal to AB; now things which are equal to the same are equal to one another, (1. Axiom); therefore CA is equal to CB; wherefore CA, AB, CB are equal to one another; and the triangle ABC is therefore equilateral, and it is described upon the given straight line AB. Which was required to be done.

PROP. II. PROB.

From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line; it is required to draw, from the point A, a straight line equal to BC.

From the point A to B draw (1. Post.) the straight line AB; and upon it describe (1 1. the equilateral triangle DAB, and produce (2. Post.) the straight lines DA, BD, to E and F; from the centre B, at the distance BC, describe (3. Post.) the circle CGH, and from the centre D, at the distance DG, describe the circle GKL. AL is equal to BC.

Because the point B is the centre of the circle CGH, BC is equal (11. Def.) to BG; and because D is the centre of the circle GKL, DL is equal to DG, and DA, DB, parts of them, are equal; therefore

C

F

K

H

D

the remainder AL is equal to the remainder (3. Ax.) BG: But it has been shewn that BC is equal to BG; wherefore AL and BC are each

of them equal to BG; and things that are equal to the same are equal to one another; therefore the straight line AL is equal to BC. Wherefore, from the given point A, a straight line AL has been drawn equal to the given straight line BC. Which was to be done.

PROP. III. PROB.

From the greater of two given straight lines to cut off a part equal to the less.

Let AB and C be the two given straight lines, whereof AB is the greater. It is required to cut off from AB, the greater, a part equal to C, the less.

From the point A draw (2. 1.) the straight line AD equal to C; and from the centre A, and at the distance AD, describe (8. Post) the circle DEF; and because A is the centre of the circle DEF, AE is

A

D

E

B

equal to AD; but the straight line C is likewise equal to AD; whence AE and C are each of them equal to AD; wherefore the straight line AE is equal to (1. Ax.) C, and from AB the greater of two straight lines, a part AE has been cut off equal to C the less. Which was to be done.

PROP. IV. THEOREM.

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If two triangles have two sides of the one equal to two sides of the other, each to each; and have likewise the angles contained by those sides equal to one another, their bases, or third sides, shall be equal; and the areas of the triangles shall be equal; and their other angles shall be equal, each to each, viz. those to which the equal sides are opposite.*

Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each, viz. AB to DE, and AC to DF; and let

the angle BAC be also equal to the angle ED F: then shall the base BC be equal to the base EF; and the triangle ABC to the triangle DEF; and the other angles, to which

A

D

the equal sides are op

The three conclusions in this enunciation are more briefly expressed by saying, that the

triangles are every way equal.

posite, shall be equal, each to each, viz. the angle ABC to the angle DEF, and the angle ACB to DFE.

For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on D, and the straight line AB upon DE; the point B shall coincide with the point E, because AB is equal to DE; and AB coinciding with DE, AC shall coincide with DF, because the angle BAC is equal to the angle EDF; wherefore also the point C shall coincide with the point F because AC is equal to DF: But the point B coincides with the point E; wherefore the base BC shall coincide with the base EF (cor. def. 3.), and shall be equal to it. Therefore also the whole triangle ABC shall coincide with the whole triangle DEF, so that the spaces which they contain or their areas are equal: and the remaining angles of the one shall coincide with the remaining angles of the other, and be equal to them, viz. the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Therefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angles contained by those sides equal to one another; their bases shall be equal, and their areas shall be equal, and their other angles, to which the equal sides are opposite, shall be equal, each to each. Which was to be demonstrated.

PROP. V. THEOR.

The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced, the angles upon the other side of the base shall also be equal.

Let ABC be an isosceles triangle, of which the side AB is equal to AC, and let the straight lines AB, AC be produced to D and E, the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE.

In BD take any point F, and from AE the greater cut off AG equal (3.1.) to AF, the less, and join FC, GB.

A

Because AF is equal to AG, and AB to AC, the two sides FA, AC are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles, AFC, AGB; therefore the base FC is equal (4. 1.) to the base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one are equal (4.1.) to the remaining angles of the other, each to each, to which the equal sides are opposite, viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB: And because the whole AF is equal to the whole AG, and the part AB to the part AC: the remainder BF shall be equal (3. Ax.) to the

F

C

B

G

D/

E

remainder CG; and FC was proved to be equal to GB, therefore the two sides BF, FC are equal to the two CG, GB, each to each; but the angle BFC is equal to the angle CGB; wherefore the triangles BFC, CGB are equal (3: 1.), and their remaining angles are equal, to which the equal sides are opposite; therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. Now, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, and the part CBG, to the part BCF the remaining angle ABC is therefore equal to the remaining angle ACB, which are the angles at the base of the triangle ABC: And it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore, the angles at the base, &c. Q. E. D.

COROLLARY. Hence every equilateral triangle is also equiangular. PROP. VI. THEOR.

İf two angles of a triangle be equal to one another, the sides which subtend, or are opposite to them, are also equal to one another.

Let ABC be a triangle having the angle ABC equal to the angle ACB; the side AB is also equal to the side AC.

D

A

For, if AB be not equal to AC, one of them is greater than the other: Let AB be the greater, and from it cut (3. 1) off DB equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC are equal to the two AC, CB.each to each; but the angle DBC is also equal to the angle ACB; therefore the base DC is equal to the base AB, and the area of the triangle DBC is equal to that of the triangle (4.1.)ACB, the less to the greater; which is absurd. Therefore, AB is not unequal to AC, that is, it is equal to it. Wherefore, if two angles, &c. Q. E. D. COR. Hence every equiangular triangle is also equilateral. PROP. VII. THEOR.

B

Upon the same base, and on the same side of it, there cannot be two triangles, that have their sides which are terminated in one extremity of the 'base equal to one another, and likewise those which are terminated in the other extremity, equal to one another.

Let there be two triangles ACB, ADB, upon the same base AB, and upon the same side of it, which have their sides CA, DA, terminated in A equal to one another; then their sides, CB, DB, terminated in

D

B, cannot be equal to one another.

Join CD, and if possible let CB be equal to DB; then, in the ease in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal (5. 1.) to the angle ADC: But the angle ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle

BCD. Again, because CB is equal to DB, the angle BDC is equal (5.1) to the angle BCD; but it has been demonstrated to be greater than it; which is impossible..

But if one of the vertices, as D, be within the other triangle ACB; produce AC, AD to E, F; therefore, because AC is equal to AD in the triangle ACD, the angies ECD, FDC upon the other side of the base CD are equal (5. 1.) to one another, but the angle ECD is greater A than the angle BCD; wherefore

the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC is equal (5. 1.) to the angle BCD; but BDC. has been proved to be greater than the same BCD; which is impossiThe case in which the vertex of one triangle is upon a side of the other, needs no demonstration.

ble.

Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise those which are terminated in the other extremity equal to one another. Q. E. D.

+ PROP. VIII. THEOR.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one shall be equal to the angle contained by the two sides of the other.

Let ABC, DEF be two triangles having the two sides AB, AC, equal to the two sides DE, DF, each to each, viz. AB to DE, and A€