#### BOOK III.

#### DEFINITIONS.

#### 1

**Equal circles**are those the diameters of which are equal, or the radii of which are equal.

#### 2

A straight line is said to**touch a circle**which, meeting the circle and being produced, does not cut the circle.

#### 3

**Circles**are said to

**touch one another**which, meeting one another, do not cut one another.

#### 4

In a circle straight lines are said**to be equally distant from the centre**when the perpendiculars drawn to them from the centre are equal.

#### 5

And that straight line is said to be**at a greater distance**on which the greater perpendicular falls.

#### 6

A**segment of a circle**is the figure contained by a straight line and a circumference of a circle.

#### 7

An**angle of a segment**is that contained by a straight line and a circumference of a circle.

#### 8

An**angle in a segment**is the angle which, when a point is taken on the circumference of the segment and straight lines are joined from it to the extremities of the straight line which is the

**base of the segment**, is contained by the straight lines so joined.

#### 9

And, when the straight lines containing the angle cut off a circumference, the angle is said to**stand upon**that circumference.

#### 10

A sector of a circle is the figure which, when an angle is constructed at the centre of the circle, is contained by the straight lines containing the angle and the circumference cut off by them.

#### 11

**Similar segments of circles**are those which admit equal angles, or in which the angles are equal to one another.

#### BOOK III. PROPOSITIONS.

#### PROPOSITION 1.

**To find the centre of a given circle**.

Let **ABC** be the given circle; thus it is required to find the centre of the circle **ABC**.

Let a straight line **AB** be drawn

through it at random, and let it be bisected at the point **D**; from **D** let **DC** be drawn at right angles to **AB** and let it be drawn through to **E**; let **CE** be bisected at **F**;

I say that **F** is the centre of the circle **ABC**.

For suppose it is not, but, if possible, let **G** be the centre, and let **GA**, **GD**, **GB** be joined.

Then, since **AD** is equal to **DB**, and **DG** is common,

**GA**is equal to the base

**GB**, for they are

radii;

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]

But the angle **FDB** is also right; therefore the angle **FDB** is equal to the angle **GDB**, the greater to the less: which is impossible.

Similarly we can prove that neither is any other point except **F**.

#### PORISM.

From this it is manifest that, if in a circle a straight line cut a straight line into two equal parts and atright angles, the centre of the circle is on the cutting straight line. Q. E. F.

#### PROPOSITION 2.

**If on the circumference of a circle two points be taken at random**,

**the straight line joining the points will fall within the circle**.

Let **ABC** be a circle, and let two points **A**, **B** be taken at random on its circumference; I say that the straight line joined from **A** to **B** will fall within the circle.

For suppose it does not, but, if possible, let it fall outside, as **AEB**; let the centre of the circle **ABC** be taken [III. 1], and let it be **D**; let **DA**, **DB** be joined, and let **DFE** be drawn through.

Then, since **DA** is equal to **DB**,

**AEB**of the triangle

**DAE**is produced,

But the angle **DAE** is equal to the angle **DBE**;

Therefore the straight line joined from **A** to **B** will not fall outside the circle.

Similarly we can prove that neither will it fall on the circumference itself;

Therefore etc. Q. E. D.

#### PROPOSITION 3.

**If in a circle a straight line through the centre bisect a straight line not through the centre**,

**it also cuts it at right angles; and if it cut it at right angles**,

**it also bisects it**.

Let **ABC** be a circle, and in it let a straight line **CD**

through the centre bisect a straight line **AB** not through the centre at the point **F**; I say that it also cuts it at right angles.

For let the centre of the circle **ABC**

be taken, and let it be **E**; let **EA**, **EB** be joined.

Then, since **AF** is equal to **FB**, and **FE** is common,

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right; [I. Def. 10]

Therefore **CD**, which is through the centre, and bisects **AB** which is not through the centre, also cuts it at right angles.

Again, let **CD** cut **AB** at right angles;

I say that it also bisects it. that is, that **AF** is equal to **FB**.

For, with the same construction,

**EAF**is also equal to the angle

**EBF**. [I. 5]

But the right angle **AFE** is equal to the right angle **BFE**,

therefore **EAF**, **EBF** are two triangles having two angles equal to two angles and one side equal to one side, namely **EF**, which is common to them, and subtends one of the equal angles;

Therefore etc. Q. E. D.
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#### PROPOSITION 4.

**If in a circle two straight lines cut one another which are not through the centre**,

**they do not bisect one another**.

Let **ABCD** be a circle, and in it let the two straight lines **AC**, **BD**, which are not through the centre, cut one another at **E**; I say that they do not bisect one another.

For, if possible, let them bisect one another, so that **AE** is equal to **EC**, and **BE** to **ED**; let the centre of the circle **ABCD** be taken [III. 1], and let it be **F**; let **FE** be joined.

Then, since a straight line **FE** through the centre bisects a straight line **AC** not through the centre,

Again, since a straight line **FE** bisects a straight line **BD**,

But the angle **FEA** was also proved right;

Therefore **AC**, **BD** do not bisect one another.

Therefore etc. Q. E. D.

#### PROPOSITION 5.

**If two circles cut one another**,

**they will not have the same centre**.

For let the circles **ABC**, **CDG** cut one another at the points **B**, **C**; I say that they will not have the same centre.

For, if possible, let it be **E**; let **EC** be joined, and let **EFG** be drawn through at random.

Then, since the point **E** is the centre of the circle **ABC**,

Again, since the point **E** is the centre of the circle **CDG**,

But **EC** was proved equal to **EF** also;

Therefore the point **E** is not the centre of the circles **ABC**, **CDG**.

Therefore etc. Q. E. D.

#### PROPOSITION 6.

**If two circles touch one another**,

**they will not have the same centre**.

For let the two circles **ABC**, **CDE** touch one another at the point **C**; I say that they will not have the same centre.

For, if possible, let it be **F**; let **FC** be joined, and let **FEB** be drawn through at random.

Then, since the point **F** is the centre of the circle **ABC**,

Again, since the point **F** is the centre of the circle **CDE**,

But **FC** was proved equal to **FB**;

Therefore **F** is not the centre of the circles **ABC**, **CDE**.

Therefore etc. Q. E. D.

#### PROPOSITION 7.

**If on the diameter of a circle a point be taken which is not the centre of the circle**,

**and from the point straight lines fall upon the circle**,

**that will be greatest on which the centre is**,

**the remainder of the same diameter will be least**,

**and of the rest**

**the nearer to the straight line through the centre is always greater than the more remote**,

**and only two equal straight lines will fall from the point on the circle**,

**one on each side of the least straight line**.

Let **ABCD** be a circle, and let **AD** be a diameter of it;

on **AD** let a point **F** be taken which is not the centre of the circle, let **E** be the centre of the circle, and from **F** let straight lines **FB**, **FC**, **FG** fall upon the circle **ABCD**; I say that **FA** is greatest, **FD** is least, and of the rest **FB** is

greater than **FC**, and **FC** than **FG**.

For let **BE**, **CE**, **GE** be joined.

Then, since in any triangle two sides are greater than the remaining one, [I. 20]

But **AE** is equal to **BE**;

Again, since **BE** is equal to **CE**, and **FE** is common,

the two sides **BE**, **EF** are equal to the two sides **CE**, **EF**.

But the angle **BEF** is also greater than the angle **CEF**; therefore the base **BF** is greater than the base **CF**. [I. 24]

For the same reason

Again, since **GF**, **FE** are greater than **EG**, and **EG** is equal to **ED**,

Let **EF** be subtracted from each;

Therefore **FA** is greatest, **FD** is least, and **FB** is greater than **FC**, and **FC** than **FG**.

I say also that from the point **F** only two equal straight lines will fall on the circle **ABCD**, one on each side of the

least **FD**.

For on the straight line **EF**, and at the point **E** on it, let the angle **FEH** be constructed equal to the angle **GEF** [I. 23], and let **FH** be joined.

Then, since **GE** is equal to **EH**,

and **EF** is common,

I say again that another straight line equal to **FG** will not

fall on the circle from the point **F**.

For, if possible, let **FK** so fall.

Then, since **FK** is equal to **FG**, and **FH** to **FG**,

Therefore another straight line equal to **GF** will not fall from the point **F** upon the circle;

Therefore etc. Q. E. D.
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#### PROPOSITION 8.

**If a point be taken outside a circle and from the point straight lines be drawn through to the circle**,

**one of which is through the centre and the others are drawn at random**,

**then**,

**of the straight lines which fall on the concave circumference**,

**that through the centre is greatest**,

**while of the rest**

**the nearer to that through the centre is always greater than the more remote**,

**but**,

**of the straight lines falling on the convex circumference**,

**that between the point and the diameter is least**,

**while of the rest the nearer to the least is always less than the more remote**,

**and only two equal straight lines will fall on the circle from the point**,

**one on each side of the least**.

Let **ABC** be a circle, and let a point **D** be taken outside **ABC**; let there be drawn through from it straight lines **DA**, **DE**, **DF**, **DC**, and let **DA** be through the centre; I say that, of the straight lines falling on the concave circumference **AEFC**, the straight line **DA** through the centre is greatest, while **DE** is greater than **DF** and **DF** than **DC**.; but, of the straight lines falling on the convex circumference **HLKG**, the straight line **DG** between the point and the diameter **AG** is least; and the nearer to the least **DG** is always less than the more remote, namely **DK** than **DL**, and **DL** than **DH**.

For let the centre of the circle **ABC** be taken [III. 1], and let it be **M**; let **ME**, **MF**, **MC**, **MK**, **ML**, **MH** be joined.

Then, since **AM** is equal to **EM**, let **MD** be added to each;

But **EM**, **MD** are greater than **ED**; [I. 20]

Again, since **ME** is equal to **MF**,

**EM**,

**MD**are equal to

**FM**,

**MD**;

Similarly we can prove that **FD** is greater than **CD**; therefore **DA** is greatest, while **DE** is greater than **DF**, and **DF** than **DC**.

Next, since **MK**, **KD** are greater than **MD**, [I. 20] and **MG** is equal to **MK**, therefore the remainder **KD** is greater than the remainder **GD**,

And, since on **MD**, one of the sides of the triangle **MLD**, two straight lines **MK**, **KD** were constructed meeting within the triangle, therefore **MK**, **KD** are less than **ML**, **LD**; [I. 21] and **MK** is equal to **ML**;

Similarly we can prove that **DL** is also less than **DH**;

I say also that only two equal straight lines will fall from the point **D** on the circle, one on each side of the least **DG**.

On the straight line **MD**, and at the point **M** on it, let the angle **DMB** be constructed equal to the angle **KMD**, and let **DB** be joined.

Then, since **MK** is equal to **MB**, and **MD** is common,

**KMD**is equal to the angle

**BMD**;

I say that no other straight line equal to the straight line **DK** will fall on the circle from the point **D**.

For, if possible, let a straight line so fall, and let it be **DN**.

while **DK** is equal to **DB**,

**DG**equal to the more remote: which was proved impossible.

Therefore no more than two equal straight lines will fall on the circle **ABC** from the point **D**, one on each side of **DG** the least.

Therefore etc.

#### PROPOSITION 9.

**If a point be taken within a circle**,

**and more than two equal straight lines fall from the point on the circle**,

**the point taken is the centre of the circle**.

Let **ABC** be a circle and **D** a point within it, and from **D** let more than two equal straight lines, namely **DA**, **DB**, **DC**, fall on the circle **ABC**; I say that the point **D** is the centre of the circle **ABC**.

For let **AB**, **BC** be joined and bisected at the points **E**, **F**, and let **ED**, **FD** be joined and drawn through to the points **G**, **K**, **H**, **L**.

Then, since **AE** is equal to **EB**, and **ED** is common,

**DA**is equal to the base

**DB**;

Therefore each of the angles **AED**, **BED** is right; [I. Def. 10] therefore **GK** cuts **AB** into two equal parts and at right angles.

And since, if in a circle a straight line cut a straight line into two equal parts and at right angles, the centre of the circle is on the cutting straight line, [