EUCLID.

BOOK I.

PROP. I. PROBLEM.

To describe an equilateral triangle upon a given

finite straight line.

Let AB be the given straight line.

It is required to describe on AB an equilateral triangle.

CONSTRUCTION.-1. From centre A, with distance AB, describe the circle BCD (post. 3).

2. From centre B, with distance BA, describe the circle ACE.

3. From point C, where the circles cut each other, draw CA and CB (post. 1).

Then, it is to be proved that

ABC is an equilateral triangle described upon AB. PROOF.-Because A is the centre of the circle BCD, therefore AB AC (def. 15).

Similarly, because B is the centre of the circle ACE, therefore BA=BC.

But it has been proved that BA=AC, therefore AC=BC (ax. 1), and therefore AB, BC, and CA = each other. Therefore, it is proved, as required, that

ABC is an equilateral triangle, described upon AB.

Q. E. F.

B

PROP. II. PROBLEM.

From a given point to draw a straight line equal to a given straight line.

Let A be the given point, and BC the given straight line.
It is required to draw from A a straight line equal to BC.

CONSTRUCTION.-1. From A to B draw the straight line AB (post. 1).

2. Upon AB describe the equilateral triangle BDA (I. 1). 3. Produce DA and DB to the points E and F (post. 2). 4. From centre B, with distance BC, describe the circle CHG, cutting DF in G (post. 3).

5. From centre D, with distance DG, describe the circle GKL cutting DE in L.

Then, it is to be proved that

AL is the line drawn from A=BC.

PROOF.-Because B is the

therefore BG=BC (def. 15).

centre of the circle CHG,

Similarly, because D is the centre of the circle GKL, therefore DG-DL.

But in the lines DG and DL we have DB=DA (cons.), therefore BG=AL (ax. 3).

Also it has been shown that BG=BC; therefore AL=BC (ax. 1).

Therefore, it is proved, as required, that

AL is the line drawn from A=BC.

Q. E. F.

PROP. III. PROBLEM.

From the greater of two given straight lines to cut off a part equal to the less.

Let AB and C be the two given straight lines, of which AB is the greater.

It is required to cut off from AB, the greater, a part equal to C, the less.

CONSTRUCTION.-1. From A draw AD=C (I. 2).

2. From centre A, with distance AD, describe the circle DEF, cutting AB in E (post. 3).

Then, it is to be proved that

AE is cut off from AB=C.

PROOF. Because A is the centre of the circle DEF, therefore AE AD (def. 15).

But AD=C (cons.); therefore AE=C (ax. 1).
Therefore, it is proved, as required, that

AE is cut off from AB=C.

Exercises.

Q. E. F.

1. Describe an equilateral triangle on a given straight line MN, with the vertex, O, below MN.

2. Prove Prop. II. when A is joined to C, instead of to B. N.B.-The Exercises given in the following pages are not necessarily connected with the Proposition under which they are placed. But they are strictly confined to that or to previous Propositions.

PROP. IV. THEOREM.

If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to each other, then they shall have their bases, or third sides, equal; and the two triangles shall be equal; and their other angles shall be equal, each to each, viz., those to which the equal sides are opposite.

In the triangles ABC and DEF let the sides AB and AC, and their angle BAC, in the former the sides DE and DF, and their angle EDF, in the latter, each to each.

PROOF. If the triangle ABC be placed upon the triangle DEF so that the point A is on the point D, and the side AB on the side DE, then, because AB=DE (hyp.),.therefore the point B shall coincide with the point E, and the side AB shall coincide with the side DE.

Next, because the side AB coincides with the side DE, and because the angle BAC = the angle EDF (hyp.), therefore the side AC shall fall on the side DF, and, because the point A coincides with the point D, and AC=DF, (hyp.), therefore the point C shall coincide with the point F.