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but the angle ECD is greater than the angle BCD; (ax. 9)

wherefore the angle FDC must likewise be greater than the angle BCD; much more then must the angle BDC be greater than the angle BCD.

Again, because in the triangle BCD, CB is assumed to be equal to DB,

therefore the angle BDC must be equal to the angle BCD ; (I. 5) but BDC has been proved to be greater than BCD;

wherefore the angle BDC must be both equal to, and greater than, the

angle BCD; which is impossible.

Thirdly. Let the vertex of one triangle be upon a side of the other.

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This case needs no demonstration, for it is evident from inspection that

the side AC is greater than its part AD. 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 ;

then 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.

(References - Prop. I. 7; ax. 8.) 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 BC equal to the base EF.

Then the angle BAC shall be equal to the angle EDF.

D

B

DEMONSTRATION

For, if the triangle ABC be applied to DEF,
so that the point B be on E, and the straight line BC upon EF ;
then because BC is equal to EF, (hyp.)

therefore the point C shall coincide with the point F; wherefore BC coinciding with EF,

BA and AC shall coincide with ED and DF; for if the base BC coincides 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

which are 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 ED, DF; wherefore the angle BAC coincides with the angle EDT,

and is equal to it. (ax. 8.) 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.

(References - Prop. I. 1, 3, 8.) Let the angle BAC be the given rectilineal angle. It is required to bisect it.

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Take any point D in AB,
from AC cut off AE equal to AD, (1. 3) and join DE;

npon the side of it opposite to A, describe the equilateral triangle DEF,

(1. 1) and join AF.
Then the straight line AF bisects the angle BAC.

DEMONSTRATION Because AD is equal to AE, (constr.) and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each; and the base DF is equal to the base EF; (constr.)

therefore the angle DAF is equal to the angle EAF. (1. 8.) Wherefore the given rectilineal angle BAC is bisected by the line AF.

Q. E. F

PROP. X.- PROBLEM.

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

(References — Prop. L 1, 4, 9.) Let AB be the given straight line. It is required to divide AB into two equal parts.

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Upon the straight line AB describe the equilateral triangle ABC, (1. 1) and bisect the angle ACB by the straight line CD. (1. 9.) Then AB shall be divided into two equal parts in the point D.

DEMONSTRATION Because AC is equal to CB, (constr.) and CD common to the two triangles ACD, BCD, the two sides AC, CD, are equal to the two sides BC, CD, each to

each ;

and the angle ACD is equal to the angle BCD; (constr.)

therefore the base AD is equal to the base DB. (I. 4.) Wherefore the straight line AB is divided into two equal parts in the point D.

Q. E. F.

PROP. XI. - PROBLEM.

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

(References — Prop. I. 1, 3, 8; ax. 1; def. 10.) Let AB be the given straight line, and C the given point in it.

It is required draw a straight line from the point C at right angles to AB.

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In AC, take any point D, and make CE equal to CD; (1. 3) apon DE describe the equilateral triangle DFE, (1. 1) and join FC. Then FC, drawn from the point C, shall be at right

angles to AB.

DEMONSTRATION

Because DC is equal to CE,
and CF is 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; (constr.)

therefore the angle DCF is equal to the angle ECF; (1. 8) and they are adjacent angles. But, 'when the adjacent angles which one straight line makes with an

other straight line, are equal to one another, each of them is called a right angle;' (def. 10) therefore each of the angles DCF, ECF, is a right angle. Wherefore, from the point C, in the straight line AB, FC has been drawn at right angles to AB,

Q. E. F.

Cor. By help of this problem, it may be demonstrated that two right

lines cannot have a common segment.

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

CONSTRUCTION

From the point B, draw BE at right angles to AB. (1. 11.)

DEMONSTRATION

Because ABC is a straight line,

therefore the angle CBE is equal to the angle EBA ; (def. 10) in the same manner, because ABD is assumed to be a straight line,

therefore the angle DBE must be equal to the angle EBA;

wherefore the angle DBE must be equal to the angle CBE, (ax. 1) the less to the greater, which is impossible.

Therefore 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.

(References — Prop. I. 8, 10; post. 3; def. 10, 15.) Let AB be the given straight line, which may be produced to any length both ways, and C the given point without it.

It is required to draw from C a straight line perpendicular to AB.

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