Sidebilder
PDF
ePub

1. The point C shall coincide with the point F, wherefore BC coinciding with EF,

B

E

2. BA and AC shall coincide with ED, DF; for, if the base BC coincide with the base EF, but the sides BA, AC do not coincide with the sides ED, DF, but have a different situation, as EG, GF: Then, upon the same base, and

upon

the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base, equal to one another, and likewise those sides which are terminated in the other extremity ; but this is impossible (I. 7). Therefore

If the base BC coincide with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore, likewise

3. The angle BAC coincides with the angle EDF, and is equal to it. (Ax. 8.)

Therefore if two triangles, &c. Q.E.D.

PROPOSITION 9.- Problem. To bisect a given rectilineal angle, that is, to divide it into two equal angles.

E

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

Construction. In AB take any point from AC cut off AED equal to A RE(I. 3), and join DE; on the side of DE remote from A, describe the equilateral triangle DEF (I. 1), and join AF.

Then the straight line AF shall bisect the angle BAC,

Proof. Because AD is equal to AE (constr.), and AF is common to the two triangles DAF, EAF,

1. The 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

3. The angle DAF is equal to the angle EAF (I. 8). Wherefore 4. The angle BAC is bisected by the straight line AF.

Q.E.F.

PROPOSITION 10.- 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 AB into two equal parts.

D

Construction. Upon AB describe the equilateral triangle ABC (I. 1), and bisect the angle ACB by the straight line CD meeting AB in the point D (I. 9).

Then AB shall be cut into two equal parts in the point D.

Proof. Because AC is equal to CB (constr.), and CD is 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

A

D

3. The base AD is equal to the base BD (I. 4). Wherefore

4. The straight line AB is divided into two equal parts in the point D.

Q.E.F.

PROPOSITION 11.- Problem.

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

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

A

D

с

В

E Construction. In AC take any point D, and make CE equal to CD (I. 3) ; upon DE describe the equilateral triangle DEF (I. 1), and join CF.

Then CF, drawn from the point C, shall be at right angles to AB. Proof. Because DC is equal to EC, and FC is common to the two triangles DCF, ECF,

1. The two sides DC, CF are equal to the two sides EC, CF,

each to each; and

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

3. The angle DCF is equal to the angle ECF (I. 8), and these two angles are adjacent angles. But when the two adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right angle (Def. 10); 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. Q.E.F. 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 segment AB be common to the two straight lines ABC, ABD.

E

[blocks in formation]

Construction. From the point B, draw BE at right angles to AB (I. 11). Demonstration. Then, because ABC is a straight line, therefore

1. The angle ABE is equal to the angle EBC (Def. 10). Similarly, because ABD is a straight line, therefore

2. The angle ABE is equal to the angle EBD; but the angle ABE is equal to the angle EBC; wherefore

3. The angle EBD is equal to the angle EBC (Ax. 1), the less equal to the greater angle, which is impossible. Therefore,

4. Two straight lines cannot have a common segment.

PROPOSITION 12.

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 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.

[blocks in formation]

Construction. Upon the other side of AB take any point D, and from the centre C, at the distance CD, describe the circle EGF, meeting AB, produced if necessary, in F and G (Post. 3) ; bisect FG in H(I. 10), and join CF, CH, CG.

Then the straight line CH drawn from the given point C, shall be perpendicular to the given straight line AB.

Proof. Because FH is equal to HG (constr.), and HC is common to the two triangles FHC, GHC;

1. The two sides FH, HC are equal to the two GH, HC, each

to each, and

2. The base CF is equal to the base CG (Def. 15); therefore

3. The angle FHC is equal to the angle GHC (I. 8), and these 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.

Q.E.F.

PROPOSITION 13.-Theorem.

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

Let the straight line AB make with CD upon one side of it, the angles CBA, ABD.

Then the angles CBA, ABD shall be either two right angles, or shall be together equal to two right angles.

« ForrigeFortsett »