# The Elements of Euclid for the Use of Schools and Colleges/Book XII

*BOOK XII.*

*If from the greater of two unequal magnitudes there be taken more than its half, and from, the remainder more than its half and so on, there shall at length remain a magnitude less than the smaller of the proposed magnitudes.*

Let *AB* and *C* be two unequal magnitudes, of which *AB* is, the greater: if from *AB* there be taken more than its half, and from the remainder more than its half, and so on, there shall at length remain a magnitude less than *C*.

For *C* may be multiplied so as at length to become greater than *AB*.

Let it be so multiplied, and let *DE* its multiple be greater than *AB*, and let *DE* be divided into *DF*, *FG*, *GE*, each equal to *C*.

From *AB* take *BH*, greater than its half, and from the remainder *AH* take *HK* greater than its half, and so on, until there be as many divisions in *AB* as in *DE*; and let the divisions in *AB* be *AK*, *KH*, *HB*, and the divisions in *DE* be *DF*, *FG*, *GE*.

Then, because *DE* is greater than *AB*;

and that *EG* taken from *DE* is not greater than its half;

but *BH* taken from *AB* is, greater than its half;

therefore the remainder *DG* is greater than the remainder *AH*. Again, because *DG* is greater than *AH*;

and that *GF* is not greater than the half of *DG*, but *HK* is greater than the half of *AH*;

therefore the remainder *DF* is greater than the remainder *AK*.

But *DF* is equal to *C*;

therefore *C* is greater than *AK*;

that is, *AK* is less than *C*. q.e.d.

And if only the halves be taken away, the same thing may in the same way be demonstrated.

*THEOREM*.

*Similar polygons inscribed in circles are to one another as the squares on their diameters.*

Let *ABCDE*, *FGHKL* be two circles, and in them the similar polygons *ABCDE*, *FGHKL*; and let *BM*, *GN* be the diameters of the circles: the polygon *ABCDE* shall be to the polygon *FGHKL* as the square on *BM* is to the square on *GM*.

Join *AM*, *BE*, *FN*, *GL*.

Then, because the polygons are similar, therefore the angle *BAE* is equal to the angle *GFL*, and *BA* is to *AE* as *GF* is to *FL*. [VI.*Definition* 1.

Therefore the triangle *BAE* is equiangular to the triangle *GFL*; [VI. 6.

therefore the angle *AEB* is equal to the angle *FLG*.

But the angle *AEB* is equal to the angle *AMB*, and the angle *FLG* is equal to the angle *FNG*; [III. 21.

therefore the angle *AMB* is equal to the angle *FNG*. And the angle *BAM* is equal to the angle *GFN*, for each of them is a right angle. [III. 31.
Therefore the remaining angles in the triangles *AMB*, *FNG* are equal, and the triangles are equiangular to one another;

therefore *BA* is to *BM* as *GF* is, to *GN*, [VI. 4.

and, alternately, *BA* is to *GF* as *BM* is to *GN*, [V. 16.

therefore the duplicate ratio of *BA* to *GF* is the same as the duplicate ratio of *BM* to *GN*. [V. *Definition* 10, V. 22.

But the polygon *ABCDE* is to the polygon *FGHKL* in the duplicate ratio of *BA* to *GF*; [VI. 20.

and the square on *BM* is to the square on *GN* in the duplicate ratio of *BM* to *GN*; [VI. 20.

therefore the polygon *ABCDE* is to the polygon *FGHKL* as the square on *BM* is to the square on *GN* [V. 11.

Wherefore, *similar polygons* &c. q.e.d.

*THEOREM*.

*Circles are to one another as the squares on their diameters.*

Let *ABCD*, *EFGH* be two circles, and *BD*, *FH* their diameters: the circle *ABCD* shall be to the circle *EFGH * as the square on *BD* is to the square on *FH*.

For, if not, the square on *BD* must be to the square on *FH* as the circle *ABCD* is to some space either less than the circle *EFGH*, or greater than it.

First, if possible, let it be as the circle *ABCD* is to a space *S* less than the circle *EFGH*.

In the circle *EFGH'* inscribe the square *EFGH*. [IV. 6.

This square shall be greater than half of the circle EFGH.

For the square *EFGH* is half of the square which can be formed by drawing straight lines to touch the circle at the points *E*, *F*, *G*, *H*;

and the square thus formed is greater than the circle;

therefore the square *EFGH* is greater than half of the circle.

Bisect the arcs *EF*, *FG*, *GH*, *HE* at the points *K*,*L*,*M*,*N*;

and join *EK*, *KF*, *FL*, *LG*, *GM*, *MH*, *HN*, *NE*. Then each of the triangles *EKF*, *FLG*, *GMH*, *HNE* shall be greater than half of the segment of the circle in which it stands.

For the triangle *EKF* is half of the parallelogram which can be formed by drawing a straight line to touch the circle at *K*, and parallel straight lines through *E* and *F*, and the parallelogram thus formed is greater than the segment *FEK*; therefore the triangle *EKF* is greater than half of the segment.

And similarly for the other triangles.

Therefore the sum of all these triangles is together greater than half of the sum of the segments of the circle in which they stand.

Again, bisect *EK*, *KF*, &c. and form triangles as before;

then the sum of these triangles is greater than half of the sum of the segments of the circle in which they stand. If this process be continued, and the triangles be supposed to be taken away, there will at length remain segments of circles which are together less than the excess of the circle *EFGH* above the space *S*, by the preceding Lemma.

Let then the segments *EK*, *KF*, *FL*, *LG*, *GM*, *MH*, *HN*, *NE* be those which remain, and which are together less than the excess of the circle above *S*;

therefore the rest of the circle, namely the polygon *EKFLGMHN*, is greater than the space *S*.

In the circle *ABCD* describe the polygon *AXBOCPDR* similar to the polygon *EKFLGMHN*.

Then the polygon *AXBOCPDR* is to the polygon *EKFLGMHN* as the square on *BD* is to the square on *FH*, [XII. 1.

that is, as the circle *ABCD* is to the space *S*. [*Hyp*., V. 11.

But the polvgon *AXBOCPDR* is less than the circle *ABCD* in which it is inscribed,

therefore the polygon *EKFLGMHN* is less than the space *S'; [V. 14.*
but it is also greater, as has been shewn;

which is impossible.

Therefore the square on *BD* is not to the square on *FH* as the circle *ABCD* is to any space less than the circle *EFGH*

In the same way it may be shewn that the square on *FH* is not to the square on *BD* as the circle *EFGH* is to any space less than the circle *ABCD*.

Nor is the square on *BD* to the square on *FH* as the circle *ABCD* is to any space greater than the circle *EFGH*.

For, if possible, let it be as the circle *ABCD* is to a space *T* greater than the circle *EFGH*.

Then, inversely, the square on *FH* is to the square on *BD* as the space *T* is to the circle *ABCD*.

But as the space *T* is to the circle *ABCD* so is the circle *EFGH* to some space, which must be less than the circle *ABCD*, because, by hypothesis, the space *T* is greater than the circle *EFGH*. [V. 14.

Therefore the square on *FH* is to the square on *BD* as the circle *EFGH* is to some space less than the circle *ABCD*;

which has been shewn to be impossible.

Therefore the square on *BD* is not to the square on *FH* as the circle *ABCD* is to any space greater than the circle *EFGH*.

And it has been shewn that the square on *BD* is not to the square on *FH* as the circle *ABCD* is to any space less than the circle *EFGH*.

Therefore the square on *BD* is to the square on *FH* as the circle *ABCD* is to the circle *EFGH*,

Wherefore, *circles* &c. q.e.d.