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terminated in the extremity A of the base equal to one another, and likewise their sides CB, DB, that are terminated in B.

A

Join CD. Then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, (I. 5.)

1. The angle ACD is equal to the angle ADC. But the angle ACD is greater than the angle BCD (Ax. 9.); therefore also

2. The angle ADC is greater than BCD; much more then

3. The angle BDC is greater than the angle BCD. Again, because CB is equal to DB, (I. 5.)

4. The angle BDC is equal 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, (I. 5.) the angles upon the other side of the base CD, namely,

1. The angles ECD, FDC, are equal to one another ; but the angle ECD is greater than the angle BCD; wherefore likewise

. 2. The angle FDC is greater than BCD; much more then

3. The angle BDC is greater than the angle BCD. Again, because CB is equal to DB, (I. 5.)

4. The angle BDC is equal 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 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.

PROP. VIII.-THEOREM. · 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 equal to them, 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 AC to DF; and also the base equal to the base EF. The angle BAC is equal to the angle EDF. A

For if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC upon EF;

1. The point C shall coincide with the point F, because BC is equal to EF; therefore BC coinciding with EF,

2. BA and AC shall coincide with ED and DF; for, if the base BC coincide with the base EF, but the sides BA, CA, do not coincide with the sides ED, FD, but have a different situation, as EG, FG; then, upon the same base EF, 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 (I. 7.); therefore, if the base BC coincide with the base EF, the sides BA, ÁC, cannot but coincide with the sides ED, DF; wherefore likewise

3. The angle BAC coincides with the angle EDF, and is equal (Ax. 8.) to it. Therefore if two triangles, &c. Q.E.D).

PROP. IX.- PROBLEM. 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 AB, and from XC cut (I. 3.) off AE equal to AD; join DE, and upon it, on the side remote from A, describe (I. 1.) an equilateral triangle DEF; then join AF; the straight line AF bisects the triangle BAC.

Because AD is equal to AE, and AF is common to the two triangles DAF, EAF;

1. T'he two sides DA, AF, are equal to the two sides EA, AF, each to each;

and (Constr.)

2. The base DF is equal to the base EF; therefore (I. 8.)

3. The angle DAF is equal to the angle EAF; wherefore

4. The angle BAC is bisected by the straight line AF. Which was to be done.

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

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

Describe (I. 1.) upon it an equilateral triangle ABC, and bisect (I. 9.) the angle ACB by the straight line CD. AB is cut into two equal parts in the point D.

A D _ Because AC is equal to CB, and CD common to the two triangles ACD, BCD;

1. The two sides AC, CD are equal to BC, CD, each to each; and (Constr.)

2. The angle ACD is equal to the angle BCD; therefore (I. 4.)

3. The base AD is equal to the base DB; wherefore

4. The straight line AB is divided into two equal parts in the point D. Which was to be done.

PROP. XI.- PROBLEM. 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 AC, and (I. 3.) make CE equal to CD, and upon DE describe (I. 1.) the equilateral triangle DFĒ, and join FC; the straight line FC drawn from the given point C is at right angles to the given straight line AB

A D C E B Because DC is equal to CE, and FC common to the two triangles DCF, ECF;

1. The two sides DC, CF, are equal to the two EC, CF, each to each; and (Constr.)

2. The base DF is equal to the base EF; therefore (I. 8.)

3. The angle DCF is equal 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 (Def. 10.) a right angle; therefore,

4. Each of the angles DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB,

5. 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.

If it be possible, let the two straight lines ABC, ABD, have the segment AB common to both of them.

From the point B draw BE at right angles to AB; and becatise ABC is a straight line, (Def. 10.)

1. The angle CBE is equal to the angle EBA; in the same manner, because ABD is a straight line,

2. The angle DBE is equal to the angle EB4; wherefore (Ax. 1.)

3. The angle DBE is equal to the angle CBE, the less to the greater, which is impossible; therefore

4. Two straight lines cannot have a common segment,

PROP. XII. -PROBLEM. 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 point without it. It is required to draw a straight line perpendicular to AB from the point C..

Take any point Dupon the other side of AB, and from the centre C, at the distance CD, describe (Post. 3.) the circle EGF meeting AB in F, G; and bisect (I. 10.) FG in H, and join CH; the straight line CH, drawn from the given point C, is perpendicular to the given straight line AB.

[blocks in formation]

Join CF, CG; and because FH is equal to HG, and HC common to the two triangles FHC, GHC,

1. The two sides FH, HC, are equal to the two GH, HC, each to each; and (Def. 15.)

2. The base CF is equal to the base CG; therefore (I. 8.)

3. The angle CHF is equal to the angle CHG; and they are adjacent angles; but when a straight line standing on another straight line makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it (Def. 10.); therefore from the given point C.

4. A perpendicular CH has been drawn to the given straight line AB. Which was to be done.

PROP. XIII.-THEOREM.

The angles which one straight line makes with another upon one side of it, are either tuo right angles, or are together equal to two right angles.

Let the straight line AB make with CD, upon one side of it, the ar:gles

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