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D

CAN DA.

СР = DB.

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Proposition 7.—Theorem. 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.

Let the triangles ACB, ADB, upon the same base AB,

and on the same side of it, have, if possible, Suppose

Their sides CĂ, DA, terminated in the extremity A of the base, equal to one another;

And their sides CB, DB, terminated in the extremity B of the base, likewiso equal to one another.

CASE I.-Let the vertex of each triangle

B be without the other triangle.
CONSTRUCTION.—Join CD.
PROOF.-Because AC is equal to AD (Hyp.),

The triangle ADC is an isosceles triangle, and the angle LACD = ACD is therefore equal to the angle ADC (I. 5).

But the angle ACD is greater than the angle BCD (Ax. 9).
Therefore the angle ADC is also greater than BCD.
Much more then is the angle BDC greater than BCD,
Again, because BC is equal to BD (Hyp.),

The triangle BCD is an isosceles triangle, and the angle
BDC is equal to the angle BCD (I. 5).

But the angle BDC has been shown to be greater than the angle BCD (Dem. 5).

Therefore the angle BDC is both equal to, and greater than the same angle BCD, which is impossible.

Case II.-Let the vertex of one of the triangles fall within the other.

CONSTRUCTION.- Produce AC, AD to E and F, and join CD.

Proof.—Because AC is equal to AD

(Hyp), Again

The triangle ADCisan isosceles triangle, B

and the angles ECD, FDC, upon the other side of its base CD, are equal to one another (I. 5),

LADC.

BDC>
BCD.

BDC = 2 BCD

L BDC = and > 2 BCD.

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A

LECD = 4 FDC.

BDO
BCD.

BCD.

But the angle ECD is greater than the angle BCD (Ax. 9). Therefore the angle FDC is likewise greater than BCD. Much more then is the angle BDC greater than BCD. Again, because BC is equal to BD (Hyp:), The triangle BDC is an isosceles triangle, and the angle 4BDC = BDC is equal to the angle BCD (I. 5).

But the angle BDC has been shown to be greater than the angle BCD.

Therefore the angle BDC is both equal to, and greater than : ZBDC the same angle BCD, which is impossible.

Therefore, upon the same base, &c. Q. E, D.

= and
> L BCD.

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Given

Proposition 8.-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. Or,

If two triangles have three sides of the one respectively equal to the three sides of the other, they are equal in every respect, those angles being equal which are opposite to the equal sides.

Let ABC, DEF be two triangles which have
The two sides AB, AC equal to the two sides DE, DF,

AB = DE, each to each, viz., AB to DE, and AC to DF,

AC = DF, And the base BC equal to the base EF. The angle BAC shall be equal to the angle EDF.

Proof.-For if the triangle ABC be applied to the triangle DEF,

So that the point B may be on E, and the straight line BC on EF,

The point C shall coincide with the point F, because BC is equal to EF (Hyp.).

Therefore, BC coinciding with EF, BA and AC shall coincide with ED B and DF.

For if the base BC coincides with the base EF,

and
BC = EF.

Make BC coincide with EF.

CE

B

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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 on the same side of it, there will be two triangles, which have their sides 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 (I. 7). .. BA, AC

Therefore, if the base BC coincides with the base EF, the respectively coincide sides BA, AC must coincide with the sides ED, DF.

Therefore the angle BAC coincides with the angle EDF, ED, DF.

and is equal to it (Ax. 8).

Also the triangle ABC coincides with the triangle DEF and is therefore equal to it in every respect (Ax. 8).

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

with

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7

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

equal parts.
Let BAC be the given rectilineal angle.

It is required to bisect it.

CONSTRUCTION—Take any point D in AB. Make

From AC cut off AE equal to AD (I. 3). AE= AD,

Join DE. A DEF e.

Upon DE, on the side remote from A, dequilateral.

scribe an equilateral triangle DEF (I. 1).
B

Join AF.
Then the straight line AF shall bisect the angle BAC.

PROOF.—Because AD is equal to AE (Const.), 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 (Const.);
Therefore the angle DAF is equal to the angle EAF (I. 8).

Therefore the given rectilineal angle BAC is bisected by the straight line AF. Q. E. F.

F

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Proposition 10.

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

Hake A ABC equilateral and L ACD = BCD.

D

B

AD

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

CONSTRUCTION.—Upon AB describe the equilateral triangle ABC (I. 1).

Bisect the angle ACB by the straight line CD (I. 9).

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

PROOF.—Because AC is equal to CB A (Const.), 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 (Const.) ;
Therefore the base AD is equal to the base DB (I. 4).

DB. Therefore 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 night 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.

CONSTRUCTION.—Take any point D in AC.
Make CE equal to CD (I. 3).

= CD and Upon DE describe the equilateral triangle DFE (I. 1). Join FC.

quilateral. Then FC shall be at right angles to AB.

PROOF.—Because DC is equal to CE (Const.), and FC common to the two triangles DCF, ECF;

The two sides DC, CF, are equal A to the two sides EC, CF, cach to each ;

And the base DF is equal to the base EF (Const.);
Therefore the angle DCF is equal to the angle ECF (I. 8); 4DCF =

) 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 the angles is called a right angle (Def. 10);

Make CE

A DEF e

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E

13

L ECF.

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IE

Make
LABE a

D

B

a

2 СТЕ LEBA.

and LDBE = LEBA,

:: L DCF,

Therefore each of the angles DCF, ECF is a right angle. L ECF are right Therefore from the given point C in the given straight line angles.

AB, a straight line FC has been drawn at right angles to
AB. Q. E. F.

COROLLARY.—By help of this problem, it may be demonstrated that

I'wo 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.

CONSTRUCTION.—From the point right 4

B, draw BE at right angles to AB

(I. 11).

с PROOF. - Because ABC is straight line, the angle CBE is equal to the angle EBA (Def. 10).

Also, because ABD is a straight line, the angle DBE is

equal to the angle EBA (Def. 10). ... DBE Therefore the angle DBE is equal to the angle CBE. The * CBE. less to the greater; which is impossible.

Therefore two straight lines cannot have a common seg. ment.

Proposition 12.- Problem. To draw a straight line perpendicular to a given straight line of 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 from the point C, a straight line perpendicular to AB.

CONSTRUCTION.—Take any point D upon

the other side of AB. From the centre C, at the distance 1 CD, describe the circle EGF, meet

ing AB in F and G (Post. 3). G B

Bisect FG in H (I. 10).

Join CF, CH, CG.
Then CH shall be perpendicular to AB.

Proof.—Because FH is equal to HG Const.), and HC cominon to the two triangles FHC, GHC.

CD as radius,

A F

Bisect FG in II.

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