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.

If it be possible, 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 the extremity A of the base c equal to one another, and likewise their sides CB, DB, and are terminated in B.

Join CD; then, in the case in which the vertex of each of the triangles is without the other tri- A

'D

B

angle, because AC is equal to AD, the angle ACD is equal (1. 5.) to the angle ACD: But the angle A CD is greater than the angle BCD (Ax. 9.); 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 (1. 5.) to the angle BCD; but it has been demonstrated to be greater than it; which is impossible.

E

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 angles ECD, FDC upon the other side of the base CD are equal (1. 5.) to one another, but the angle ECD is greater than the angle BCD; wherefore the angle FDC is likewise greater than BCD; much more

A

D

B

then is the angle B D C greater than the angle BCD. Again, because C B is equal to D B, the angle B DC is equal (1. 5.) to the angle BCD; but BDC has been proved to be greater than the same BCD, which is impossible. The case in which the vertex of the triangle is upon a side of the other, needs no demonstration.

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. 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 sides equal to them, of the other.

Let A B C, DEF be two triangles having the two sides A B, A C, equal to the two sides, DE, DF, each to each, viz. AB to DE, and A

AC to DF, and also the base BC equal to the base E F. The angle BAC is equal to the angle E DF.

B

D G

CE

For if the triangle ABC be applied to DEF, so that the point в be on E, and the straight line BC upon EF; the point C shall also coincide with the point BC because BC is equal to EF (Hyp.); therefore BC coinciding with EF, BA and AC shall coincide with ED and DF; for, if the base B C coincides with

C

the base E F, but the sides BA, CA do not coincide with the sides E D, F D, but have a different situation as E G, F G; then, upon the same base E F, and upon the same side of it, there can be two triangles that have their sides which are terminated in one extremity of the base equal to one another, and likewise their sides terminated in the other extremity: But this is impossible (1. 7.); therefore, if the base BC coincides with the base EF; the sides BA, AC cannot but coincide with the sides E D, D F, wherefore likewise the angle BAC coincides with the angle E D F, and is equal (Ax. 8.) to it. Therefore if two triangles, &c. Q. E. D.

PROP. IX. PROB.

To bisect a given rectilineal angle, that is, to divide it into two equal angles.

Let BAC be the given rectilineal angle, it is required to bisect it.

Take any point D in A B, and from AC cut (1. 3.) off A E equal to A D ; join D E, and upon it, on the side towards B and C, describe (1. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the triangle BA C.

Because A D is equal to A E, and AF is common to the two triangles DAF, EAF; the two sides DA, A F, are equal to the two sides E A, AF, each to each; and the base DF is equal to the base E F; therefore the B

D

A

E

C

angle DAF is equal (1. 8.) to the angle EA F; wherefore the given rectilineal angle BAC is bisected by the straight line a F. Which was to be done.

PROP. X. PROB.

To bisect a given finite straight line, that is, to divide it into two equal parts.

Let A B be the given straight line; it is required to divide it into two equal parts.

Describe (1. 1.) upon it an equilateral triangle A B C, and bisect (1. 9.) the angle A C B by the straight AB is cut into two equal parts in the

line C D.

point D.

C

Because AC is equal to C B, and CD common to the two triangles ACD, BCD; the two sides A C, CD are equal to BC, CD, each to each: and the angle ACD is equal to the angle BCD (Constr.); therefore the base A D is equal to the base (1. 4.) D B, and the straight line AB is divided into two equal parts in the point D.

A

D

B

Which was to be done.

PROP. XI. PROB.

To draw a straight line at right angles to a given straight line, from a given point in the same.

Let AB be a given straight line, and c a point given in it; it is required to draw a straight line from the point c at right angles to AB.

Take any point D in AD, and (1. 3.) make CE equal to CD, and upon DE des

F

cribe (1. 1.) the equilateral tri

angle DFE, and join FC; the

straight line FC drawn from
the given point c is at right
angles to the given straight A D

line AB.

Because DC is equal to CE, and FC common to

the two triangles DCF, ECF; the two sides DC; CF, are equal to the two EC, CF, each to each; and the base DF is equal to the base EF; therefore the angle DCF is equal (1. 8.) to the angle ECF; and they are adjacent angles: but, when the adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right (1. Def. 10.) angle; therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the given point c, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done.

COR. By help of this problem, it may be demonstrated that two straight lines cannot have a common segment.

E

If it be possible, let the two straight lines A B C, ABD, have the segment A B common to both of them. From the point в draw BE at right angles to AB; and because ABC is a straight line, the angle CBE is equal (1. Def. 10.) to the angle EBA; in the same manner, because A B D is a straight line, the angle DBE is equal to the angle E BA; wherefore the angle DBE is equal to the angle CBE, the less to the greater, which is ☎

B

D

C

impossible; therefore two straight lines cannot have a common segment.

PROP. XII. PROB.

To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it.

Let AB be the given straight line, which may be produced to any length both ways, and let c be a