#### BOOK II. PROPOSITIONS.

#### Proposition 1.

**If there be two straight lines**,

**and one of them be cut into any number of segments whatever**,

**the rectangle contained by the two straight lines is equal to the rectangles contained by the uncut straight line and each of the segments**.

Let **A**, **BC** be two straight lines, and let **BC** be cut at random at the points **D**, **E**; I say that the rectangle contained by **A**, **BC** is equal to the rectangle contained by **A**, **BD**, that contained by **A**, **DE** and

that contained by **A**, **EC**.

For let **BF** be drawn from **B** at right angles to **BC**; [I. 11] let **BG** be made equal to **A**, [I. 3] through **G** let **GH** be drawn

parallel to **BC**, [I. 31] and through **D**, **E**, **C** let **DK**, **EL**, **CH** be drawn parallel to **BG**.

Then **BH** is equal to **BK**, **DL**, **EH**.

Now **BH** is the rectangle **A**, **BC**, for it is contained by **GB**, **BC**, and **BG** is equal to **A**;

**BK** is the rectangle **A**, **BD**, for it is contained by **GB**, **BD**, and **BG** is equal to **A**;

Similarly also **EH** is the rectangle **A**, **EC**.

Therefore the rectangle **A**, **BC** is equal to the rectangle **A**, **BD**, the rectangle **A**, **DE** and the rectangle **A**, **EC**.

Therefore etc. Q. E. D.
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#### Proposition 2.

**If a straight line be cut at random**,

**the rectangle contained by the whole and both of the segments is equal to the square on the whole**.

For let the straight line **AB** be cut at random at the point **C**; I say that the rectangle contained by **AB**, **BC** together with the rectangle contained by **BA**, **AC** is equal to the square on **AB**.

For let the square **ADEB** be described on **AB** [I. 46], and let **CF** be drawn through **C** parallel to either **AD** or **BE**. [I. 31]

Then **AE** is equal to **AF**, **CE**.

Now **AE** is the square on **AB**;

**AF** is the rectangle contained by **BA**, **AC**, for it is contained by **DA**, **AC**, and **AD** is equal to **AB**;

Therefore the rectangle **BA**, **AC** together with the rectangle **AB**, **BC** is equal to the square on **AB**.

Therefore etc. Q. E. D.

#### Proposition 3.

**If a straight line be cut at random**,

**the rectangle contained by the whole and one of the segments is equal to the rectangle contained by the segments and the square on the aforesaid segment**.

For let the straight line **AB** be cut at random at **C**; I say that the rectangle contained by **AB**, **BC** is equal to the rectangle contained by **AC**, **CB** together with the square on **BC**.

For let the square **CDEB** be described on **CB**; [I. 46] let **ED** be drawn through to **F**, and through **A** let **AF** be drawn parallel to either **CD** or **BE**. [I. 31]

Then **AE** is equal to **AD**, **CE**.

Now **AE** is the rectangle contained by **AB**, **BC**, for it is contained by **AB**, **BE**, and **BE** is equal to **BC**;

**AD** is the rectangle **AC**, **CB**, for **DC** is equal to **CB**;

**AB**,

**BC**is equal to the rectangle contained by

**AC**,

**CB**together with the square on

**BC**.

Therefore etc. Q. E. D.

#### Proposition 4.

**If a straight line be cut at random**,

**the square on the whole is equal to the squares on the segments and twice the rectangle contained by the segments**.

For let the straight line **AB** be cut at random at **C**;

I say that the square on **AB** is equal to the squares on **AC**, **CB** and twice the rectangle contained by **AC**, **CB**.

For let the square **ADEB** be described on **AB**, [I. 46]

let **BD** be joined; through **C** let **CF** be drawn parallel to either **AD** or **EB**, and through **G** let **HK** be drawn parallel to either **AB** or **DE**. [I. 31]

Then, since **CF** is parallel to **AD**, and **BD** has fallen on them, the exterior angle **CGB** is equal to the interior and opposite angle **ADB**. [I. 29]

But the angle **ADB** is equal to the angle **ABD**,

**CGB**is also equal to the angle

**GBC**, so that the side

**BC**is also equal to the side

**CG**. [I. 6]

But **CB** is equal to **GK**, and **CG** to **KB**; [I. 34]

I say next that it is also right-angled.

For, since **CG** is parallel to **BK**,

But the angle **KBC** is right;

Therefore **CGKB** is right-angled; and it was also proved equilateral;

For the same reason

Therefore the squares **HF**, **KC** are the squares on **AC**, **CB**.

Now, since **AG** is equal to **GE**, and **AG** is the rectangle **AC**, **CB**, for **GC** is equal to **CB**,

Therefore **AG**, **GE** are equal to twice the rectangle **AC**, **CB**.

But the squares **HF**, **CK** are also the squares on **AC**, **CB**; therefore the four areas **HF**, **CK**, **AG**, **GE** are equal to the squares on **AC**, **CB** and twice the rectangle contained by **AC**, **CB**.

But **HF**, **CK**, **AG**, **GE** are the whole **ADEB**,

which is the square on **AB**.

Therefore the square on **AB** is equal to the squares on **AC**, **CB** and twice the rectangle contained by **AC**, **CB**.

Therefore etc. Q. E. D.
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#### Proposition 5.

**If a straight line be cut into equal and unequal segments, the rectangle contained by the unequal segments of the whole together with the square on the straight line between the points of section is equal to the square on the half**.

For let a straight line **AB** be cut into equal segments at **C** and into unequal segments at **D**; I say that the rectangle contained by **AD**, **DB** together with the square on **CD** is equal to the square on **CB**.

For let the square **CEFB** be described on **CB**, [I. 46] and let **BE** be joined; through **D** let **DG** be drawn parallel to either **CE** or **BF**, through **H** again let **KM** be drawn parallel to either **AB** or **EF**, and again through **A** let **AK** be drawn parallel to either **CL** or **BM**. [I. 31]

Then, since the complement **CH** is equal to the complement **HF**, [I. 43] let **DM** be added to each;

But **CM** is equal to **AL**,

**CH**be added to each;

But **AH** is the rectangle **AD**, **DB**, for **DH** is equal to **DB**,

Let **LG**, which is equal to the square on **CD**, be added to each;

But the gnomon **NOP** and **LG** are the whole square **CEFB**, which is described on **CB**;

Therefore etc. Q. E. D.
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#### Proposition 6.

**If a straight line be bisected and a straight line be added to it in a straight line**,

**the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line**.

For let a straight line **AB** be bisected at the point **C**, and let a straight line **BD** be added to it in a straight line;

I say that the rectangle contained by **AD**, **DB** together with the square on **CB** is equal to the square on **CD**.

For let the square **CEFD** be described on **CD**, [I. 46] and let **DE** be joined; through the point **B** let **BG** be drawn parallel to either **EC** or **DF**, through the point **H** let **KM** be drawn parallel to either **AB** or **EF**, and further through **A** let **AK** be drawn parallel to either **CL** or **DM**. [I. 31]

**CH**is equal to

**HF**. [I. 43]

Let **CM** be added to each;

But **AM** is the rectangle **AD**, **DB**,

Let **LG**, which is equal to the square on **BC**, be added to each;

But the gnomon **NOP** and **LG** are the whole square **CEFD**, which is described on **CD**;

Therefore etc. Q. E. D.

#### Proposition 7.

**If a straight line be cut at random**,

**the square on the whole and that on one of the segments both together are equal to twice the rectangle contained by the whole and the said segment and the square on the remaining segment**.

For let a straight line **AB** be cut at random at the point **C**;

I say that the squares on **AB**, **BC** are equal to twice the rectangle contained by **AB**, **BC** and the square on **CA**.

For let the square **ADEB** be described on **AB**, [I. 46] and let the figure be drawn.

Then, since **AG** is equal to **GE**, [I. 43] let **CF** be added to each;

Therefore **AF**, **CE** are double of **AF**.

But **AF**, **CE** are the gnomon **KLM** and the square **CF**; therefore the gnomon **KLM** and the square **CF** are double of **AF**.

But twice the rectangle **AB**, **BC** is also double of **AF**; for **BF** is equal to **BC**; therefore the gnomon **KLM** and the square **CF** are equal to twice the rectangle **AB**, **BC**.

Let **DG**, which is the square on **AC**, be added to each; therefore the gnomon **KLM** and the squares **BG**, **GD** are equal to twice the rectangle contained by **AB**, **BC** and the square on **AC**.

But the gnomon **KLM** and the squares **BG**, **GD** are the whole **ADEB** and **CF**,

**AB**,

**BC**are equal to twice the rectangle contained by

**AB**,

**BC**together with the square on

**AC**.

Therefore etc. Q. E. D.

#### Proposition 8.

**If a straight line be cut at random**,

**four times the rectangle contained by the whole and one of the segments together with the square on the remaining segment is equal to the square described on the whole and the aforesaid segment as on one straight line**.

For let a straight line **AB** be cut at random at the point **C**;

I say that four times the rectangle contained by **AB**, **BC** together with the square on **AC** is equal to the square described on **AB**, **BC** as on one straight line.

For let [the straight line] **BD** be produced in a straight line [with **AB**], and let **BD** be made equal to **CB**; let the square **AEFD** be described on **AD**, and let the figure be drawn double.

Then, since **CB** is equal to **BD**, while **CB** is equal to **GK**, and **BD** to **KN**, therefore **GK** is also equal to **KN**.

For the same reason

And, since **BC** is equal to **BD**, and **GK** to **KN**, therefore **CK** is also equal to **KD**, and **GR** to **RN**. [I. 36]

But **CK** is equal to **RN**, for they are complements of the parallelogram **CP**; [I. 43] therefore **KD** is also equal to **GR**; therefore the four areas **DK**, **CK**, **GR**, **RN** are equal to one another.

Again, since **CB** is equal to **BD**, while **BD** is equal to **BK**, that is **CG**, and **CB** is equal to **GK**, that is **GQ**,

And, since **CG** is equal to **GQ**, and **QR** to **RP**,

But **MQ** is equal to **QL**, for they are complements of the parallelogram **ML**; [I. 43]

**AG**,

**MQ**,

**QL**,

**RF**are equal to one another.

**CK**,

**KD**,

**GR**,

**RN**were proved to be quadruple of

**CK**;

Now, since **AK** is the rectangle **AB**, **BD**, for **BK** is equal to **BD**, therefore four times the rectangle **AB**, **BD** is quadruple of **AK**.

But the gnomon **STU** was also proved to be quadruple of **AK**;

Let **OH**, which is equal to the square on **AC**, be added to each;

But the gnomon **STU** and **OH** are the whole square **AEFD**,

**AB**,

**BD**together with the square on

**AC**is equal to the square on

**AD**

But **BD** is equal to **BC**; therefore four times the rectangle contained by **AB**, **BC** together with the square on **AC** is equal to the square on **AD**, that is to the square described on **AB** and **BC** as on one straight line.

Therefore etc. Q. E. D.

#### Proposition 9.

**If a straight line be cut into equal and unequal segments**,

**the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section**.

For let a straight line **AB** be cut into equal segments at **C**, and into unequal segments at **D**;

I say that the squares on **AD**, **DB** are double of the squares on **AC**, **CD**.

For let **CE** be drawn from **C** at right angles to **AB**, and let it be made equal to either **AC** or **CB**; let **EA**, **EB** be joined, let **DF** be drawn through **D** parallel to **EC**, and **FG** through **F** parallel to **AB**, and let **AF** be joined.

Then, since **AC** is equal to **CE**,

And, since the angle at **C** is right,

And they are equal;

For the same reason

And, since the angle **GEF** is half a right angle, and the angle **EGF** is right, for it is equal to the interior and opposite angle **ECB**, [I. 29]

Again, since the angle at **B** is half a right angle, and the angle **FDB** is right, for it is again equal to the interior and opposite angle **ECB**, [I. 29]

Now, since **AC** is equal to **CE**,

But the square on **EA** is equal to the squares on **AC**, **CE**, for the angle **ACE** is right; [I. 47]

Again, since **EG** is equal to **GF**,

But the square on **EF** is equal to the squares on **EG**, **GF**;

But **GF** is equal to **CD**; [I. 34] therefore the square on **EF** is double of the square on **CD**.

But the square on **EA** is also double of the square on **AC**;

And the square on **AF** is equal to the squares on **AE**, **EF**, for the angle **AEF** is right; [I. 47] therefore the square on **AF** is double of the squares on **AC**, **CD**.

But the squares on **AD**, **DF** are equal to the square on **AF**, for the angle at **D** is right; [I. 47] therefore the squares on **AD**, **DF** are double of the squares on **AC**, **CD**.

And **DF** is equal to **DB**; therefore the squares on **AD**, **DB** are double of the squares on **AC**, **CD**.

Therefore etc. Q. E. D.

#### Proposition 10.

**If a straight line be bisected**,

**and a straight line be added to it in a straight line**,

**the square on the whole with the added straight line and the square on the added straight line both together are double of the square on the half and of the square described on the straight line made up of the half and the added straight line as on one straight line**.

For let a straight line **AB** be bisected at **C**, and let a straight line **BD** be added to it in a straight line;

I say that the squares on **AD**, **DB** are double of the squares on **AC**, **CD**.

For let **CE** be drawn from the point **C** at right angles to **AB** [I. 11], and let it be made equal to either **AC** or **CB** [I. 3]; let **EA**, **EB** be joined; through **E** let **EF** be drawn parallel to **AD**, and through **D** let **FD** be drawn parallel to **CE**. [I. 31]

Then, since a straight line **EF** falls on the parallel straight lines **EC**, **FD**,

But straight lines produced from angles less than two right angles meet; [I. Post. 5]

Let them be produced and meet at **G**, and let **AG** be joined.

Then, since **AC** is equal to **CE**, the angle **EAC** is also equal to the angle **AEC**; [I. 5] and the angle at **C** is right;

For the same reason

And, since the angle **EBC** is half a right angle, the angle **DBG** is also half a right angle. [I. 15]

**DCE**, they being alternate; [I. 29]

**DGB**is equal to the angle

**DBG**,

Again, since the angle **EGF** is half a right angle, and the angle at **F** is right, for it is equal to the opposite angle, the angle at **C**, [I. 34]

Now, since the square on **EC** is equal to the square on **CA**, the squares on **EC**, **CA** are double of the square on **CA**.

But the square on **EA** is equal to the squares on **EC**, **CA**; [I. 47] therefore the square on **EA** is double of the square on **AC**. [C. N. 1]

Again, since **FG** is equal to **EF**, the square on **FG** is also equal to the square on **FE**; therefore the squares on **GF**, **FE** are double of the square on **EF**.

But the square on **EG** is equal to the squares on **GF**, **FE**; [I. 47] therefore the square on **EG** is double of the square on **EF**.

And **EF** is equal to **CD**; [I. 34]

**EA**was also proved double of the square on

**AC**; therefore the squares on

**AE**,

**EG**are double of the squares on

**AC**,

**CD**.

And the square on **AG** is equal to the squares on **AE**, **EG**; [I. 47] therefore the square on **AG** is double of the squares on **AC**, **CD**. But the squares on **AD**, **DG** are equal to the square on **AG**; [I. 47] therefore the squares on **AD**, **DG** are double of the squares on **AC**, **CD**.

And **DG** is equal to **DB**; therefore the squares on **AD**, **DB** are double of the squares on **AC**, **CD**.

Therefore etc. Q. E. D.

#### Proposition 11.

**To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment**.

Let **AB** be the given straight line; thus it is required to cut **AB** so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

For let the square **ABDC** be described on **AB**; [I. 46] let **AC** be bisected at the point **E**, and let **BE** be joined; let **CA** be drawn through to **F**, and let **EF** be made equal to **BE**; let the square **FH** be described on **AF**, and let **GH** be drawn through to **K**.

I say that **AB** has been cut at **H** so as to make the rectangle contained by **AB**, **BH** equal to the square on **AH**.

For, since the straight line **AC** has been bisected at **E**, and **FA** is added to it,

But **EF** is equal to **EB**;

But the squares on **BA**, **AE** are equal to the square on **EB**, for the angle at **A** is right; [I. 47]

Let the square on **AE** be subtracted from each;

Now the rectangle **CF**, **FA** is **FK**, for **AF** is equal to **FG**; and the square on **AB** is **AD**;

Let **AK** be subtracted from each;

And **HD** is the rectangle **AB**, **BH**, for **AB** is equal to **BD**; and **FH** is the square on **AH**;

#### Proposition 12.

**In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle**,

**namely that on which the**

**perpendicular falls**,

**and the straight line cut off outside by the perpendicular towards the obtuse angle**.

Let **ABC** be an obtuse-angled triangle having the angle **BAC** obtuse, and let **BD** be drawn from the point **B** perpendicular to **CA** produced;

I say that the square on **BC** is greater than the squares on **BA**, **AC** by twice the rectangle contained by **CA**, **AD**.

For, since the straight line **CD** has been cut at random at the point **A**, the square on **DC** is equal to the squares on **CA**, **AD** and twice the rectangle contained by **CA**, **AD**. [II. 4]

Let the square on **DB** be added to each; therefore the squares on **CD**, **DB** are equal to the squares on **CA**, **AD**, **DB** and twice the rectangle **CA**, **AD**.

But the square on **CB** is equal to the squares on **CD**, **DB**, for the angle at **D** is right; [I. 47]

**CB**is equal to the squares on

**CA**,

**AB**and twice the rectangle contained by

**CA**,

**AD**;

Therefore etc. Q. E. D.

#### Proposition 13.

**In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle**,

**namely that on which the perpendicular falls**,

**and the straight line cut off within by the perpendicular towards the acutc angle**.

Let **ABC** be an acute-angled triangle having the angle at **B** acute, and let **AD** be drawn from the point **A** perpendicular to **BC**;

I say that the square on **AC** is less than the squares on **CB**, **BA** by twice the rectangle contained by **CB**, **BD**.

For, since the straight line **CB** has been cut at random at **D**,

Let the square on **DA** be added to each; therefore the squares on **CB**, **BD**, **DA** are equal to twice the rectangle contained by **CB**, **BD** and the squares on **AD**, **DC**.

But the square on **AB** is equal to the squares on **BD**, **DA**, for the angle at **D** is right; [I. 47] and the square on **AC** is equal to the squares on **AD**, **DC**; therefore the squares on **CB**, **BA** are equal to the square on **AC** and twice the rectangle **CB**, **BD**,

so that the square on **AC** alone is less than the squares on **CB**, **BA** by twice the rectangle contained by **CB**, **BD**.

Therefore etc. Q. E. D.

#### Proposition 14.

**To construct a square equal to a given rectilineal figure**.

Let **A** be the given rectilineal figure; thus it is required to construct a square equal to the rectilineal figure **A**.

For let there be constructed the rectangular parallelogram **BD** equal to the rectilineal figure **A**. [I. 45]

Then, if **BE** is equal to **ED**, that which was enjoined will have been done; for a square **BD** has been constructed equal to the rectilineal figure **A**.

But, if not, one of the straight lines **BE**, **ED** is greater.

Let **BE** be greater, and let it be produced to **F**; let **EF** be made equal to **ED**, and let **BF** be bisected at **G**.

With centre **G** and distance one of the straight lines **GB**, **GF** let the semicircle **BHF** be described; let **DE** be produced

to **H**, and let **GH** be joined.

Then, since the straight line **BF** has been cut into equal segments at **G**, and into unequal segments at **E**,

But **GF** is equal to **GH**; therefore the rectangle **BE**, **EF** together with the square on **GE** is equal to the square on **GH**.

But the squares on **HE**, **EG** are equal to the square on **GH**; [I. 47]

therefore the rectangle **BE**, **EF** together with the square on **GE** is equal to the squares on **HE**, **EG**.

Let the square on **GE** be subtracted from each;

But the rectangle **BE**, **EF** is **BD**, for **EF** is equal to **ED**;

And **BD** is equal to the rectilineal figure **A**.

Therefore the rectilineal figure **A** is also equal to the square

which can be described on **EH**.

Therefore a square, namely that which can be described on **EH**, has been constructed equal to the given rectilineal figure **A**. Q. E. F.