PROPOSITION VI.

If two angles and the adjacent side of one triangle be respectively equal to two angles and the adjacent side of another triangle, then the triangles shall be equal in all respects.

да

In the triangles ABC, DEF, let the two angles ABC, ACB, and the adjacent side BC, be respectively equal to the two angles DEF, DFE, and the adjacent side EF. Then shall the triangles ABC DEF be equal in all respects.

For if the triangle ABC were applied to the triangle DEF, so that BC fell upon EF,

then BA would fall along ED, since the angle ABC is equal to the angle DEF;

and CA would fall along FD, since the angle ACB is equal to the angle DFE.

Therefore the point A would fall on the point D, and the triangle ABC would coincide with the triangle DEF Therefore the triangles ABC, DEF are equal in all respects.

Wherefore if two angles and the adjacent side &c.

Q.E.D.

PROPOSITION VII.

If the three sides of one triangle be respectively equal to the three sides of another, the triangles shall be equal in all respects.

Let the sides AB, BC, CA of the A ABC be respectively equal to the sides DE, EF, FD of the ▲ DEF Then shall the triangles ABC, DEF be equal in all respects.

For if the ▲ ABC were taken up, reversed, and placed so that BC fell on EF,

opposite side of EF to that on which

EHD be equal to ▲ EDH,

and A on the

then would

because EH is equal to ED;

(I. 5)

and

FHD be equal to

FDH,

because FH is equal to FD;

(I. 5)

and therefore L EHF equal to

EDF,

that is,

BAC is equal to ▲ EDF,

also the sides AB, AC are respectively equal to the

sides DE, DF.

Therefore the ▲ ABC is equal to the ▲ DEF in all respects.

Wherefore if three sides of one triangle &c.

(I. 4)

Q.E.D.

Note. For convenience, the side BC of A ABC, which is not less than either of the others, has been applied to the equal side of A DEF.

PROPOSITION VIII.

To bisect a given angle.

Let DAE be the given angle.

It is required to bisect it.

With centre A describe a circle cutting AD, AE B and C.

With centres B and C describe circles having equ radii, intersecting in F.

Join AF.

Then DAE shall be bisected by AF.

Join BF and CF.

Because in the As ABF, ACF,

the side AF is common to both;

also BA is equal to CA, because they are radii of t same circle,

and BF is equal to CF, because they are radii equal circles;

therefore the As ABF, ACF are equal in all

spects,

and thus

Wherefore

BAF is equal to the CAF
DAE has been bisected.

(I.

Q.E

With centres A and B describe two circles having equal radii intersecting in C and D.

Join CD cutting AB in E.

Then AB shall be bisected in E.

Join AC, CB, BD, DA.

Now because in the As A CD, BCD the sides AC, CD, DA are respectively equal to BC, CD, DB,

therefore the As ACD, BCD are equal in all re

spects,

ACD is equal to the

BCD.

and thus the (I. 7) Hence in the As ACE, BCE, the two sides AC, CE and the included angle ACE are respectively equal to the two sides BC, CE and the included angle BCE. Therefore the As ACE, BCE are equal in all respects, and therefore AE is equal to BE. (I. 4) Wherefore the given straight line AB has been bisected.

Q.E.F.

PROPOSITION X.

At a point in a given straight line to make an angle equal to a given angle.

AA

Let BAC be the given angle.

It is required to make at the point D in the straight line DE an angle equal the given angle BAC.

With centre A describe a circle cutting AB and AC in Fand G; and with centre D describe a circle HKM with equal radius cutting DE in H.

Join FG; and with centre H and radius equal to FG describe a circle cutting the circle HKM in K.

Join DK.

Then EDK shall be the angle required.

Join HK.

Then because in the As HDK, FAG the sides HD, DK, and KH are respectively equal to FA, AG, and GF;

therefore HDK is equal to LFAG.

(I. 7)

Wherefore at the point D in the straight line DE an angle EDK has been made equal to the given angle ВАС.

Q.E. F.