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PROP. I. PROBLEM.

A.

B

E

To describe un equilateral triangle upon a given finite straight line. Let AB be the given straight line. It is required to describe an equilateral triangle upon AB.

From the centre Ā, at the distance AB, describe (Post. 3) the circle BCD. 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 (Post. 1) CA, CB, to the points A, B. Then ABC is an equilateral triangle.

Because the point A is the centre of the circle BCD, AC is equal (Def. 15) to AB. And because the point B is the centre of the circle ACE, B'C is equal to BA. But it has been proved that CA is equal to AB. Therefore the two straight lines CA, CB, are each of them equal to AB. But things which are equal to the same thing are equal (Ax. 1) to one another. Therefore CA is equal to CB. Wherefore the three sides CA, AB, BC, are equal to one another. The triangle ABC is, therefore, equilateral. And it is described upon the given straight line AB. Q. E. F.

From the construction of this problem, it is plain that, upon any given straight line,

two equilateral triangles may be described, viz. one on each side. Ecercise. To describe an isosceles triangle upon a given finite straight line, that

shall have each of its sides double the base.

PROP. II. PROBLEM. 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 (Post. 1) the straight line AB. Upon A B describe (I. 1) the equilateral triangle DAB. And produce (Post. 2) the straight lines DA, DB, to the points E and F. From the centre B, at the distance BC, describe (Post. 3) the circle CGH. And from the centre D, at the distance D G, describe the circle G KL. Then, the straight line AL is equal to B C.

Because the point B is the centre of the circle CGH, B C is equal (Def. 15) to BG. And because the point D is the centre of the circle GKL, DL is equal to D G. But (Const.) D A, D B, parts of these equals, are equal. Therefore the remainder AL is equal to the remainder (Ax. 3) BG.. But it has been shown that B C is equal to BG. Wherefore AL and BC are each of them equal to BG.

And things that are equal to the same thing are equal (Ax. 1) to one another. Therefore the straight line AL is equal to the straight line BC. Wherefore from the given point A, a straight line AL has been drawn equal to the given straight line B C. Q. E. F.

The construction of this problem might be improved thus:- Join AB. Upon AB

describe the equilateral triangle ABD. From the centre B, at the distance

K

D

B

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CI

BC, describe the circle CGH. Produce DB to meet the circumference in G.
From the centre D, at the distance DG, describe the circle G KL. Produce
D A to meet the circumference in L. Then AL is equal to B C. The demon.

stration of this construction will be the same as that above. Exercise. Draw the figures, and show the application of the construction and

demonstration to different positions of the point and the straight line; such as, when the given point is situated above the straight line or below the straight line; also, when in the straight line itself, at the extremities, or at any point between them,

PRO P. III. PROBLEM.

D

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A.

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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, of which 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 (I. 2) the straight line AD equal to C. And from the centre A, at the distance AD, describe (Post. 3) the circle DEF. Then the part AE shall be equal to C.

Because A is the centre of the circle DEF, A E is equal (Def. 15) to AD. But the straight line C is likewise equal (Const.) to AD. Therefore AE and C are each of them equal to AD. Wherefore the straight line AE is equal (Ax. 1) to Ĉ. Therefore from AB the greater of two given straight lines, a part AE has been cut off equal to C the less. Q. E. F. Exercise. To produce the smaller of two given straight lines, so that with the part

produced, it shall be equal to the greater.

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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 these sides equal to one another, their bases, or third sides, are equal ; the two triangles are equal; and their other angles are equal, each to each ; viz.,

those to which the equal sides are opposite. Let ABC, D E F be two triangles, which. have the two sides AB, AC equal to the two sides DE, DF, each to each; viz., A B to DE, and AC to DF; and the angle BAC equal to the angle EDF. Then the base BC is equal to the base EF; the triangle ABC is equal to the triangle DEF; and the remaining angles of the one are equal to the remaining angles of the other, each to each ; viz., those to which the equal sides are opposite; that is, the angle A B C is equal to the angle DEF, and the angle A CB to the angle DFE.

For, if the triangle ABC be applied to the triangle DEF, so that the point A may be on the point I), and the straight line A B upon the straight line DE. The point B shall coincide (that is, fall upon, so as to agree) with the point E, because AB is equal (Hyp.) to D E. And AB coinciding with DE, A C shall coincide with D F, because the angle

B

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BAC is equal (Hyp:) to the angle EDF. Also, the point C shall coincide with the point F, because AC (Hyp.) is equal to DF. But the point B was proved to coincide with the point E. Therefore the base B C shall coincide with the base EF. For, the point B coinciding with the point E, and the point C with the point F, if the base BC does not coincide with the base E F, the two straight lines B C, EF, would enclose a space, which (Ax. 10) is impossible. Wherefore the base BC coincides with the base EF, and is, therefore, equal (Ax. 8) to it. Wherefore, also, the whole triangle A B C coincides with the whole triangle DEF, and is, therefore, equal to it. And the remaining angles of the one coincide with the remaining angles of the other, and are therefore equal to them, each to each ; viz., the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Therefore, if two triangles have two sides, &c. Q. E. D.

The demonstration of this proposition might have been conducted by beginning

the application of the one triangle to the other at the points B and E, and

then going round the figure as above. This proposition holds equally true, when the angles contained by the two sides, of

the one triangle is the same as that contained by the two sides of the other; as, in the triangles FAC, GAB, see fig. to next proposition ; or, when the triangles have a common base, as in the triangles FBC, GCB, see the same fig. ; or when

they have a common side, Corollary. If the sides A B, D Е, or the sides AC, DF, were produced, it would

be shown in the same manner, that the angles formed upon the other sides of

the bases of the triangles would be equal, each to each. Exercise. If two squares have one side of the one equal to one side of the other, the squares are equal in all respects.

PROP. V. THEOREM. 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 be equal. Let ABC be an isosceles triangle, of which the side AB is equal to the side AC. And let the equal sides AB, AC be produced to D and E. Then, the angle ABC is equal to the angle ACB, and the angle CBD to the angle BCE.

In B D take any point F, and from A E the greater, cut off AG equal (I. 3) to AF the less. Join FC, GB.

Because in the two triangles AFC, AGB, AF is equal to (Const.) AG, and AB to (Hyp.) AC, the two sides FA, AC are equal to the two sides GA, AB, each to each ; and they contain the angle FAG common to the two triangles AFC, AGB. Therefore the base FC is equal (I. 4) to the base GB, and the

А. triangle AFC to the triangle AGB. Also, the remaining angles of the one are equal (I. 4) to the remaining angles of the other, each to each ; viz., those to which the equal sides are opposite; that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. Again, because the whole AF is equal to the whole AG, of which the parts AB, AC, are equal. Therefore the remainder BF is equal (Ax. 3) to the remainder CG. But, FC was proved to be

B

F

D

equal to GB. Therefore, in the two triangles BFC, GCB, the two sides BF, FC are equal to the two sides CG, GB, each to each. And the angle BFC was proved to be equal to the angle CG B. Wherefore the two triangles BFC, CGB, are equal (I. 4), and their remaining angles are equal, each to each, viz., those 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. But it has been demonstrated, that the whole angle ABG is equal to the whole angle ACF, and the part CBG of the one, equal to the part BCF of the other. Therefore the remaining angle ABC is equal (Ax. 3) to the remaining angle ACB, and these are the angles at the base of the triangle A B C. It has also been proved that the angle FBC is equal to the angle GCB, and these are the angles upon the other side of the base. Therefore the angles at the base, &c. Q. E. D.

Corollary. Hence every equilateral (equal-sided) triangle is also equiangular (equal.

angled). This demonstration might be shortened by the application of the corollary to the

preceding proposition. To do so, will be a useful exercise to the student. The enunciation of this proposition is more clearly expressed thus : “ If two sides of

a triangle be equal to one another, the angles which are opposite to the equal sides, are also equal to one another;" and if the equal sides be produced, the angles upon the other side of the base shall likewise be equal.

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If two angles of a triangle be equal to one another, the sides which subtend,

or are opposite to the equal angles, are equal to one another. Let ABC be a triangle having the angle A BC equal to the angle ACB. Then the side A B is equal to the side AC.

For, if AB be not equal to AC, one of them is greater than the other Let AB be the greater; and from B A cut ( 1.3) off BD equal to A C, the less. And join DC.

Because in the two triangles DBC, ACB, the side DB is equal to the side AC, and BC is common to both, the two sides, DB, BC are equal to the two sides AC, CB, each to each. And the angle DBC is equal to the angle (Hyp.) ACB. Therefore the base DC is equal to the base AB. And the triangle DBC is equal to the triangle (I. 4) ACB, the less to the greater, which is absurd. Therefore AB is not unequal to AC; that is, A B is equal to A C. Wherefore, if two angles, &c. Q. E. D.

B
Corollary. Hence every equiangular triangle is also equilateral.
This proposition is called the converse of Prop. V., because the

hypothesis of that proposition is the predicate of this one. Exercise. In the figure of Prop. V., let a straight line be drawn from the point

A to the point of intersection of the two straight lines BG. FC. It is required to prove that this straight line will bisect the angle FAG.

This suggests a readier mode of bisecting an angle than that contained in Prop. IX. of this book.

PRO P. VII. THEORE M.

D

B

Upon the same base, and on the same side of it, there cannot be two

triangles having their sides terminated in one extremity of the base, equal to one another, and likewise those terminated in the other

extremity. If it be possible, upon the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, having their sides CA, DA terminated in the extremity A of the base, equal to one another, and likewise their sides, CB, DB, terminated in the extremity B.

Join CD. First, let the vertex of each triangle be without the other triangle. Because AC is equal (Hyp.) to AD in the triangle ACD, the angle ACD is equal (I. 5) to the angle ADC. But the angle ACD is greater (Ax. 9) than the angle BCD. Therefore the angle A ADC is also greater than the angle BCD. Much more then is the angle BDC greater than the angle. BCD. Again, because CB is equal (Hyp.) to DB, the angle BDC is equal (1.5) to the angle BCD. But it has been demonstrated that the angle BDC is greater than the angle BCD. Therefore, the angle BDC is both equal to, and greater than the angle BCD; which is impossible.

Secondly, let the vertex of one of the triangles be within the other triangle. Produce AC and AD to Eand F. Because AC is equal (Hyp.) to AD in the triangle ACD, the angles: ECD, FDC upon the other side of the base CD are equal (I. 5) to one another. But the angle ECD is greater (Ax. 9) than the angle BCD. Therefore the angle FD C is likewise greater than BCD. Much more then is the angle BDC greater than the angle BCD. Again, because CB is equal (Hyp.) to DB, the angle A BDC is equal (I. 5) to the angle BCD. But it has been proved that the angle BDC is greater than the angle BCD. Therefore, the angle BDC is both equal to, and greater than the angle BCD, which is impossible.

The third case, in which the vertex of one triangle is upon a side of the other needs no demonstration.

Therefore, upon the same base, and on the same side of it, &c. Q. E. D.

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The third case needs no demonstration, because when the vertex of the one

triangle is upon a side of the other, that is, on a point between A and C, in the triangle ACB, the sides which terminate in one extremity of the base are unequal, which is contrary to the hypothesis.

Excercise. If two triangles on the same base, and on opposite sides of it, have their

sides terminated in one extremity of the base equal, and likewise those terminated in the other extremity; the angle contained by the two sides of the one shall be equal to the angle contained by the two sides of the other.

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