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For Euclid, I. 5, see Appendix to Book I.

PROP. 5. THEOR.

The angles at the base of an isosceles triangle are equal.

Given A ABC, having AB = AC. .
To prove < B = L C.
Suppose AD bisects < BAC.
:: BA = AC (given), and AD is com-

mon to the triangles BAD and CAD, and BAD = L CAD (supposition), :: A BAD

ACAD (I. 4) and : LB = L C. Q.E.D.

Note.-This proposition, if not accepted as a sufficient substitute for Euclid's, may serve to tide the beginner over the formidable Fifth until he has acquired some facility in geometrical methods and reasoning; then, say at a second reading, he will be able to master with comparative ease Euclid's Fifth, given in the Appendix. The only objection to the above is that in it we make use of a line which we have not yet learned how to find, but which it is evident does exist.

It is an easy corollary to I. 13, to prove that the angles below the base formed by producing the equal sides are also equal.

5. The line which bisects the vertical angle of an isosceles triangle bisects the base, and is at right angles to the base.

6. Every equilateral triangle is also equiangular.

7. If two isosceles triangles are on the same base and on opposite sides of it, the straight line joining the vertices bisects the vertical angles, bisects the common base, and is at right angles to the base.

8. The same is true if the isosceles triangles are on the same side of the base. (Şee note, Eu. I. 7, Appendix.)

9. If any number of isosceles triangles be on the same base, their vertices shall be all in one straight line. (This straight line is called the locus of points equidistant from the extremi. ties of the given base.)

= AC

The part

PROP. 6. THEOR. If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, those angles shall be equal to one another. Given a ABC, having - ABC

A
ACB.
To prove AB AC.
If AB be not = AC, one must be >

the other.
Let AB be > AC; cut off BD
(Post. 3). Join DC.

В.

C :: DB

AC, and BC is common to the triangles DBC and ACB, and _ DBC = ACB (given), .:. Δ DBC = A ACB (1. 4).

the whole, which is impossible; :. AB is not unequal to AC.

AC. Q.E.D. Note.-The method of proof employed in this proposition (reductio ad absurdum) usually presents difficulty to the beginner. What he ought to note is, that we prove AB=AC, by disproving what contradicts it. either = AC or not = AC. The one supposition, that AB is not = AC, leads to a false conclusion, and therefore it must be a false supposition. Therefore the other supposition, that AB=AC, must be the true one.

10. Every equiangular triangle is also equilateral.

11. If the opposite angles of a quadrilateral are equal, and the diagonals bisect each of the angles through which they pass, the figure is equilateral.

12. If the straight line joining the vertex of a triangle to the middle point of the base is also at right angles to the base, the triangle is isosceles.

13. If two straight lines bisect one another at right angles, any point in either is equidistant from the extremities of the other (Prop. 4).

14. If two quadrilaterals have three sides of the one equal respectively to three sides of the other, and likewise the angles contained by the equal sides equal in each, the figures shall be equal in every respect.

:: AB

AB is

B

For Euclid I. 7 and I. 8, see Appendix.

PROP. 8. THEOR. If two triangles have the three sides of the one respectively equal to the three sides of the other, the triangles shall be equal in every respect; and of the angles those are equal which are opposite corresponding sides.

B

E

F

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Given AB DE, AC = DF, and BC = EF.
To prove A BAC

EDF in every respect.
Place A ABC on A DEF so that B is on E, and BC

along EF; then C falls on F : BC = EF (given). And let A ABC take the position of A GEF on the opposite side of EF to A DEF. Join DG.

: EG is BA it = ED (given),
and ,: EDG = EGD (1.5).
Similarly - FDG = 2 FGD.
.: LEDF = - EGF (Ax. 2); but - EGF is A,

- EDF
Again :: BA, AC = ED, DF (given), and L A =

, _EDF (proved), :: ABAC

AEDF in every respect (I. 4). Q.E.D. 15. Prove the above proposition when DG passes through the point F, and also when DG falls beyond F.

16. The straight line joining the vertex of an isosceles triangle to the middle point of the base is at right angles to the base, and bisects the vertical angle.

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PROP. 9. PROB. To bisect a given angle.

Given < BAC.
To bisect - BAC.
Take any point D in AB and
cut off in AC, AE AD

E
(Post. 3). Join DE; and on
DE on the side remote from

B
A describe an equil. ADFE
(1. 1). Join AF. AF bisects – BAС.
: DA

AE (cons.) and AF is common, and DF FE (cons.), :: A DAF

AEAF in every respect (1. 8). ::DAF

= EAF, and .; - BAC is bisected. Q.E.F.

PROP. 10. PROB.
To bisect a given finite straight line.
Given AB a straight line.
To bisect AB.
On AB describe an equil. A ACB

(1.1).
Bisect – ACB by line CD (1.9).
AB is bisected at D.

B .: AC CB (cons.), and CD is common, and L ACD

LBCD (cons.), :: AACD

ABCD in every respect (I. 4).

BD, and .. AB is bisected. Q.E.F. 17. In Prop. 9, why should the equil. A DFE be drawn on the side remote from A ? Need A DFE be equilateral ?

18. To divide a given angle into four equal parts.

19. To divide a given finite straight line into four, eight, etc., equal parts.

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PROP. 11. PROB. To draw a straight line at right angles to a given straight line, from a given point in the same. Given st. line AB with point C in it.

F To draw from C a st.

line | AB.
With C as centre cut

off, in AB, CD and
CE equal lengths.

A D

с E B On DE describe an equil. DFE (1. 1). Join FC.

FC is I AB.
In triangles DCF and ECF,
: DC = CE (cons.), and CF is common, and FD

FE (Def. 17),
.:. Δ DCF A ECF (1.8), and LDCF

_ ECF, and these 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. :: CF is at right angles to AB, and is drawn from the

given point C. Q.E.F. Note.-Since an angle may be equal to two right angles (Def. 6), then Prop. 11 is a particular case of Prop. 9.

Special cases of this and of the next proposition may require a line to be produced, and this has been granted (Post. 2).

20. Only one straight line perpendicular to a given straight line can be drawn from a point in the same.

21. To draw a straight line at right angles to a given straight line from its middle point ; also from its extremity.

22. In a given straight line to find a point equidistant from two given points. That this is not always possible can be proved by I. 28.

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