General Enunciation.

To describe an equilateral triangle on a given finite straight line.

(A finite line is one whose length is definitely given.) Particular Enunciation.

A

Given. The line AB. Required. To describe an equilateral triangle on A B.

Construction.

(a) From the centre A,
at the distance A B,
describe the
BCD.

circle

A

B

(b) From the centre B, at the distance A B, describe the circle ACE.

D

E

B

(c) From the point C, at which the circles cut one another, draw the lines CA, CB, to A and B.

Then ACB shall be an equilateral triangle (now we must prove that A B C is equilateral).

Proof (with complete syllogisms).

Ist Syllogism.

All lines drawn from the centre of a circle to the circumference are equal (Definition of a circle).

A is the centre of a circle, and A B and A C are drawn from it to the circumference (Const.). (a) ... A B is equal to A C (1st conclusion). 2nd Syllogism.

All lines drawn from the centre of a circle to the circumference are equal (Definition of a circle).

B is the centre of a circle, and A B and B C are drawn from it to the circumference.

(b) ... A B is equal to B C (2nd conclusion). 3rd Syllogism.

Things equal to the same thing are equal to one another (Axiom 1).

A C and B C are each equal to A B (a and b).
... AC is equal to BC.

Result... AC, BC, AB are all equal to one another.

Wherefore the triangle A B C is equilateral, and it is described on the given straight line A B.

Q. E. F.

Proof (with contracted syllogisms, in the words of Euclid).

(a). A C is equal to AB.
Because B is the centre
.. BC is equal to A B.

Because A is the centre of the circle BCD.
(Definition of circle.)
of the circle A CF.
(Definition of circle.)
But it has been proved that A C is equal to
AB (a).

(b) A C and B C are each equal to A B.

Things which are equal to the same thing are equal to one another (Axiom 1).

(c). A C is equal to B C.

.. AC, BC, AB are equal to one another (b, c).

Wherefore the triangle A B C is equilateral, and it is described on the given straight line A B.

Q. E. F.

EXERCISES.-I. Describe an equilateral triangle having each of its sides double the given straight line O P.

0

P

II. Describe two equilateral triangles, one on each side of the given line I 2 and

prove that the five sides are all equal (use figures instead of letters to name the lines and points).

PROBLEM (Euclid I. 22).

Repeat.-The definitions of a circle and a triangle. And Axiom 1.

General Enunciation.

To make a triangle, of which the sides shall be equal to three given lines.

(Of the three given lines, any two must always be greater than the third.)

Particular Enunciation.

Given. The three straight lines, A, B, C, of which A and B are greater than C: A and C are greater than B: B and C are greater than A.

с

B

Required. To make a triangle, whose sides shall be equal to A B and C.

Construction.

(a) Draw a straight line DE, of unlimited length towards E.

D

F

Cut off DF equal to line
ine B; G H equal to line C.
how to do this.)

(d) Join FK, K G.

H

K

b) From Fas centre, with the radius FD, describe the circle D KL.

G

(c) From G as centre, with radius GH, describe the circle HKL, cutting the former circle at K.

A; FG equal to (Euc. I. 3 shows

H E

The triangle shall have its sides

FK equal to A; FG equal to B; G K equal to C.

We are required to prove three things:

1. That F K is equal to A.
2. That F G is equal to B.
3. That G K is equal to C.

Proof (with complete syllogisms) of No. 1.

Ist Syllogism.

All lines drawn from the centre of a circle to the circumference are equal (Definition of circle).

Fis the centre of circle DK L, and FK, FD

are drawn from it to the circumference (Const.). (a) .:. F K is equal to FD. (1st conclusion.) 2nd Syllogism.

(a) FK is equal to FD. (1st conclusion become a premiss.)

A is equal to FD (by construction).

(1).. FK is equal to A. (One of the conclusions wanted.)

Proof (with complete syllogisms) of No. 3.

Ist Syllogism.

All lines drawn from the centre of a circle to the circumference are equal. (Definition.)

G is the centre of circle HKL, and GK, G H are drawn from it to the circumference.

(a).. G H is equal to G K. (1st conclusion. 2nd Syllogism.

(a) GH is equal to GK. (1st conclusion become a premiss.)

GH is equal to C (by construction).

(3). G K is equal to C. (Axiom 1.) (3rd conclusion wanted.)

Proof of No. 2.

(2) F G was made equal to B.

Result.-Collecting together conclusions (1), (2), (3), we have:

The three sides KF, FG, G K are equal to the three lines A, B, C.