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Because in the two isosceles triangles ABC and ADF, the sides AB, AC are equal to the sides AD, AF (def. 6), and the angle BAC contained by the sides AB, AC is equal to the angle FAD (Prop. 2) contained by the sides AD, AF, therefore the triangles ABC and ADF are equal (Prop. 3), and the angles at B and C are equal to the angles at D and F. Because those angles are equal, which are opposite to the equal sides (Prop. 3); therefore the angle at B is equal to either of the angles at D and F (as the side AC, to which the angle at B is opposite, is equal to either of the sides AD and AF to which the angles at F and D are opposite). Eor the same reason the angle at C is equal to either of the angles at D and F. But things which are equal to the same thing are equal to one another, therefore the angles at B and C which are equal to the same thing D or F, are equal.

Conversely, let the angles at B and C be equal, then the sides BA and AC are equal.

Because it can be proved, that no part of AB can be equal to AC. For, if possible, let BD a part of AB be equal to AC. Join DC.

Because in the two triangles DBC, ACB, the side DB is equal to the side AC (hyp), and BC is common to both, the two sides DB, BC, are equal to the two sides AC, CB each to each. And the angle DBC is equal

D

to the angle ABC. Therefore the base DC is equal to the base AB. And the triangle DBC is equal to the triangle (P. 3) ACB, the less to the greater, which is absurd. In similar way, it can be to proved, that no part of AC could be equal to AB. Therefore AB is not unequal to AC; that is, AB is equal to AC. Wherefore, the angles at the base &c. Q.E.D.

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To bisect a given finite straight line and to draw a perpendicular to it from the point of bisection.

Let AB be a finite straight line. It is required to bisect AB and to raise a perpendicular from the point of bisection.

From the point A, as centre, with the

straight line AB, draw the circle GBH and from B with BA

draw the circle AD.

AG and AH are

Join AG, BG, AH, BH and GH. equal to AB (Def. 6). And also BG,

(Def. 6). therefore AG, AH, BG,

BH are equal to AB
BH are all equal (Ax. 1).

Because the side AG is equal to the side BG, therefore the angle GAB of the triangle AGB is equal to the angle GBA (P. 4). In like manner it can be proved, that the angle HAB is equal to the angle HBA (P. 4). Therefore the whole angle GAH is equal to the whole angle GBH (Ax. 2). Because in the two triangles AGH and BGH the sides AG, AH and BG, BH are equal and the vertical angles GAH and GBH are also equal. Therefore, the angle AGH at the base of the triangle AGH is equal to the angle BGH at the base of the triangle BGH (p. 3).

Because in the two triagles AGF and FGB, the side AG is equal to the side BG and the side GF is common and also because the vertical angle AGF is equal to the vertical angle FGB, therefore the bases AF and FB are equal (P. 3). Wherefore AB is bisected at F

Also the angle AFG is equal to the angle BFG (p. 3). But when a straight line standing on another straight line, makes the adjacent angles equal to one another, eac

of these angles is called a right angle, and the straight line which stands on the other, is called a perpendicular to it. Therefore the angles AFG and BFG are right angles and the straight line GF is drawn a perpendicular from the point F. Wherefore the straight line A B is bisected at F and FG drawn perpendicular to it from the point of bisection. Q. E. F.,

Cor. 1. It is plain that a perpendicular can be drawn from any given point in a straight line (E 1. 11).

Because, if AF be a straight line, then from F a given point in the straight line AF, FG can be drawn perpendicular to AF, by producing AF and making AF and BF equal.

Cor. 2. The angles which one straight line makes with another upon one side of it, are together equal to two right angles (E. 1. 13).

Because, if FD a straight line makes, with AB, two angles at the point F, and if FG be drawn perpendicular from F, the two angles made by FD are together equal to the two right angles made by FG (Ax. 8).

Cor. 3. To draw an equilateral triangle upon a given straight line (E. 1. 1.)

AGB is an equilateral triangle upon AB, the given straight line.

PROP. VI. THEOREM. (E. 1. 16.)

If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles,

Let ABC be a triangle, and its side BC be produced to D. The exterior angle ACD is greater than either of the rior opposite angles CBA, BAC,

Bisect (P. 5) AC in F, join BF and produce it to E. Make EF equal (Def. 6) to BF. Join EC.

'G

FB, are equal

But the angle

Because AF is equal/Const.) to FC, and BF/Const.) to EF. Therefore, in the triangles AFB, CFE, the two sides AF, to the two sides CF, EF, each to each. AFB is equal (P. 3) to the angle CFE, because they are vertical angles. Therefore the base AB is equal (P. 3) to the base CE, the triangle AFB to the triangle CFE, and the remaining angles of the one to the remaining angles of the other, each to each :-viz., those to which the equal sides are opposite. Wherefore the angle BAF is equal to the angle ECF. But the angle FCD is greater (Ax. 7) than the angle ECF. Therefore the angle ACD is greater than the angle BAF. In the same manner, if the side BC be bisected, and AC be produced to G, it may be demonstrated that the angle BCG, is greater than the angle ABC. But the angle ACD is equal (P. 2) to the angle BCG. Therefore the angle ACD is greater than the angle ABC. Therefore, if one side, &c. Q. E. D.

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If a straight line falling on two other straight lines, makes the alternate angles equal to each other; these two straight lines are parallel, and conversely, if a straight line falls upon two parallel straight lines, it makes the alternate angles equal. Let the straight line EF, which falls upon the two straight lines AB, CD, make the alternate angles AEF, EFD, equal to one another. Then AB shall be parallel to CD.

.A

C

B

For, if AB be not parallel to CD, AB and CD being produced will meet either towards A and C, or towards B and D. Let AB, CD be produced and meet towards B and D, in the point G. Then GEF is a triangle, and its exterior angle AEF is greater (P. 6) than its interior and opposite angle EFG. But the angle AEF is equal (Hyp.) to the angle EFG. Therfore the angle AEF is both greater than, and equal to the angle EFG; which is impossible. Wherefore AB, CD being produced, do not meet towards BD. In like manner, it may be prov→ ed, that they do not meet when produced towards A, C. But those straight lines in the same plane, which do not meet either way, though produced ever so far, are paral lel (Def. 23) to one another. Therefore. AB is parallel to CD.

Conversely, if AB and CD be two parallel straight lines, then EF falling upon them, makes the alternate angles AEF and EFD equal.

Because it can be proved, that no angle either less or greater than EFD can be equal to AEF. For, if possible, let the angle EFH which is less than the angle EFD be equal to the angle AEF. Because EF, a straight line, falling upon AB and FH, two other straight lines, makes the alternate angles AEF and EFH equal, therefore EH is (1st case) parallel to AB. But CD is parallel to AB (hyp). Therefore EH and CD which intersect each other are both parallel to AB. This is impossible (Ax. 10). For the same reason, no angle greater than EFD can be equal to AEF. Wherefore, if a straight line &c. Q.E.D.

PROP. VIII. THEOREM. (E. 1, 28-29.)

If a straight line fall upon two parallel straight lines; it makes the exterior angle equal to the interior and opposite

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