PROP. VI. THEOR.

In any triangle (ABC) if two angles (ABC and ACB) are equal, the sides (AB, AC) opposite to them are also equal.

A

C

Let it be assumed that the sides AB and AC, opposite to the equal angles, are not equal, and that a part DC, cut off from AC, conterminous with the third side, may be equal to AB. Draw BD; and since, in the triangles ABC, DCB, the side AB is assumed to be equal to DC, BC is com- B4 mon to both, and ▲ ABC=/ DCB (Hyp.), the triangles must be equal (Prop. 4). But the triangle DCB is part of ABC, and is therefore also less than it (Ax. 9), which is absurd. Therefore DC cannot be equal to AB: and the same may be demonstrated of any other part of AC, which is therefore not greater than AB. In like manner AB may be shown to be not greater than AC. Therefore the sides AB and AC subtending the equal angles B and C are equal.

COR.-Hence every equiangular triangle is also equilateral.

PROP. VII. THEOR.

On the same base (AB), and on the same side of it, there cannot be two triangles having their conterminous sides (AC and AD, BC and BD) at both extremities of the base, equal to each other.

When two triangles stand on the same base, and on the same side of it, the vertex of the one shall either fall outside of the other triangle, or within it; or, lastly, on one of its sides.

C D

In the first case, when the vertex of each falls outside of the other triangle, draw CD joining the vertices and then because AC=AD (Hyp.), ČAD is an isosceles triangle, and ACD=/ ADC (Prop. 5); but ACD is greater than BCD, which is but a part of it; therefore ADC also is greater than BCD. Again, since BC=BD (Hyp.), ▲ BCD =/ BDC (Prop. 5); but it has been proved that ADC is greater than BCD; it must therefore be also greater

A

B

than BDC, of which it is but a part, which is absurd: therefore the two triangles cannot have their conterminous sides equal at both extremities of the base, viz. AC to AD and BC to BD, when the vertex of the one falls outside of the other triangle.

E

C/F

B

Next, suppose the vertex of the one to fall within the other triangle; and draw, as before, CD joining the vertices, and produce a pair of the conterminous sides, AC, AD beyond CD towards E and F. Then, since AC=AD (Hyp.), ≤s ECD, FDC, below the base of the isosceles triangle CAD are equal (Prop. 5); but ECD is greater than BCD, which is but a part of it (Ax. 9), A and therefore FDC also must be greater than Again, since BD=BC (Hyp.), ≤ BDC=/ BCD (Prop. 5.): but it has been proved that FDC is greater than BCD; it must therefore be also greater than / BDC, of which however it is but a part; which is absurd. Therefore, two triangles on the same base, and on the same side of it, cannot have their conterminous sides equal at both extremities of the base, when the vertex of the one falls within the other triangle.

BCD.

A

When the vertex D of one of the triangles falls on a side CA of the other, it is evident that the conterminous sides CD and CA are unequal. Therefore in no case can there be on the same base, and on the same side of it, two triangles having their conterminous sides equal at both extremities of B the base.

PROP. VIII. THEOR.

If two triangles (ACB and DFE) have two sides (CA, CB) of the one respectively equal to two sides (FD, FE) of the other, and also their bases (AB and DE) equal; then the angles (c and F) contained by their equal sides are also equal.

Let the triangles be so applied that the extremity A of the base of the one may coincide with that extremity D of the base of the other, at which terminates the side DF equal to AC; and that AB may fall A

C

F

B

on DE, and the triangles ACB, DFE may lie at the same side. Then, since the extremities A and D of the equal bases coincide, their other extremities B and E shall also coincide, and therefore the bases themselves shall wholly coincide (Def. 3). The two triangles ACB, DFE stand therefore on the same base, and on the same side of it, and their conterminous sides being equal they must coincide, since there cannot be two triangles under such conditions (Prop. 7). But since the side AC must coincide with DF, and BC with EF, it is evident that the angle C is equal to the angle F.

[The triangles ACB, DFE therefore having two sides CA, CB in the one respectively equal to two sides FD, FE in the other, and the angles C and F contained by the equal sides also equal, are equal in every respect (Prop. 4).]

PROP. IX. PROB.

To bisect a given rectilinear angle (BAC).

A

In the leg AB of the given angle take any point D; cut off from the other leg AC the part AE equal to AD (Prop. 3); draw the straight line DE, and construct upon it, at the side remote from A, the equilateral triangle DFE, and draw the straight line AF: it shall bisect the given angle.

B

D

B

For the triangles DAF and EAF have the sides AD AE, and FD=FE (Const.), and AF is common to both; they are therefore equal, and have s DAF, EAF, which are subtended by the equal sides FD, FE, equal (Prop. 8). Therefore the line AF bisects the angle BAC.

PROP. X. PROB.

To bisect a given finite straight line (AB).

On the given line AB, construct an equilateral triangle ACB (Prop. 1), and bisect its vertical angle C (Prop. 9). The line CD bisecting the vertical angle shall also bisect the base of the equilateral triangle.

For since the triangles ACD, BCD have the

sides CA = CB, CD common to both, and s DCA, DCB contained by the equal sides, also equal (Const.); their bases AD and BD must be equal (Prop. 4). Therefore the given

line AB is bisected in D.

PROP. XI. PROB.

From a given point (c) in a given straight line (AB), to draw a perpendicular to the given line.

F

Take any point D in the given line, and in the opposite direction from C cut off CE equal to CD; upon DE construct an equilateral triangle DFE, and from its vertex to the given point draw the straight line FC. The line FC shall be perpendicular to the given line.

B

D

C

E

For the triangles DFC, EFC have the sides CD=CE, and FD=FE (Const.), and FC is common to both: they are therefore equal (Prop. 8), and the angles DCF and ECF, which are subtended by equal sides, are equal; therefore, since FC standing on AB makes the adjacent angles equal, it is perpendicular to AB (Def. 7).

PROP. XII. PROB.

To draw a straight line perpendicular to a given indefinite straight line (AB) from a given point (c) without.

C

At the side of the given line remote from C, take any point P, and from C as centre, and with the interval CP as radius, describe a circle, cutting AB in E and F. Bisect EF in D (Prop. 10), and draw CD. This line CD shall be perpendicular to AB.

A E

P

For draw CE and CF; and since the triangles CED, CFD have CD common to both, and the sides CE=CF (Def. 12), and DE=DF (Const.), they shall be equal (Prop. 8); and the angles CDE, CDF opposite to equal sides shall be equal: therefore the straight line CD standing on AB, so as to make the adjacent angles equal, is perpendicular to it (Def. 7).

PROP. XIII. THEOR.

When a straight line (AB) standing upon another straight line (CD) makes angles with it; they are either two right angles, or together equal to two right angles.

F

If AB be perpendicular to CD, then s ABC, ABD must be two right angles (Def. 7). But if AB be not perpendicular to CD, then from the point B, draw BF perpendicular to CD, and on the same side of it with BA (Prop. 11). Now s FBC, FBD are two right angles (Const.); but FBD= < FBA+ABD; and FBC being added to both, FBD + 2 FBC=/ FBC + 2 FBA + ▲ ABD (Ax. 2) ; but FBC + / FBA = 2 ABC; therefore / ABC + ABD =/ FBD+/ FBC; or the angles made with CD, by the line BA standing upon it, not perpendicularly, are equal to two right angles.

C

B Ꭰ

COR. Since the angles made at any point on one side of a straight line, are equal to two right angles; it is manifest that the angles at any point in a straight line, on both sides of it, or all the angles round a point, are together equal to four right angles.

PROP. XIV. THEOR.

If two straight lines (CB and DB), meeting a third straight line (AB), at the same point (B), and at opposite sides of it, make with it the adjacent angles equal to two right angles; these straight lines lie in one continuous straight line.

For, if it be supposed that BD is not the continuation of CB; but that CB being produced, BF is the produced part; then, since AB stands upon the line CF, ▲ ABC + 2 ABF = two right angles (Prop. 13); but also ABC+/ABD=two right angles (Hyp.); therefore / ABC+2 ABF =2 ABC+2 ABD (Ax. 1), and ▲ ABC being taken from both, there remains / ABF=▲ ABD (Ax. 3); a part equal to the whole, which is absurd (Ax. 9). Therefore BF is not the con

C

F

B

D