And because the angle ADC in a semicircle is a right angle, (III. 31.) and because in the right-angled triangle ADC, BD is drawn from the right angle perpendicular to the base,

DB is a mean proportional between AB, BC the segments of the base: (VI. 8. Cor.)

therefore between the two given straight lines AB, BC, a mean proportional DB is found.

Q.E.F.

PROPOSITION XIV. THEOREM.

Equal parallelograms, which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and conversely, parallelograms that have one angle of the one equal to one angle of the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let AB, BC be equal parallelograms, which have the angles at B equal.

The sides of the parallelograms AB, BC about the equal angles, shall be reciprocally proportional;

that is, DB shall be to BE, as GB to BF.

Let the sides DB, BE be placed in the same straight line;
wherefore also FB, BG are in one straight line: (1. 14.)
complete the parallelogram FE.

And because the parallelogram AB is equal to BC, and that FE is another parallelogram,

AB is to FE, as BC to FE: (v. 7.)

but as AB to FE so is the base DB to BE, (vI. 1.)
and as BC to FE, so is the base GB to BF;
therefore, as DB to BE, so is GB to BF. (v. 11.)

Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

Next, let the sides about the equal angles be reciprocally proportional, viz. as DB to BE, so GB to BF:

the parallelogram AB shall be equal to the parallelogram BC. Because, as DB to BE, so is GB to BF';

and as DB to BE, so is the parallelogram AB to the parallelogram FE; (VI. 1.)

and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore as AB to FE, so BC to FE: (v. 11.) therefore the parallelogram AB is equal to the parallelogram BC. (v. 9.)

Therefore equal parallelograms, &c. Q.E.D.

PROPOSITION XV. THEOREM.

Equal triangles which have one angle of the one equal to one angle of the other, have their sides about the equal angles reciprocally proportional: and conversely, triangles which have one angle in the one equal to one angle in the other, and their sides about the equal angles reciprocally proportional, are equal to one another.

Let ABC, ADE be equal triangles, which have the angle BAC equal to the angle DAE.

Then the sides about the equal angles of the triangles shall be reciprocally proportional;

that is, CA shall be to AD, as EA to AB.

Let the triangles be placed so that their sides CA, AD be in one straight line;

wherefore also EA and AB are in one straight line; (1. 14.)

and join BD.

Because the triangle ABC is equal to the triangle ADE,
and that ABD is another triangle;

therefore as the triangle CAB, is to the triangle BAD, so is the triangle AED to the triangle DAB; (v. 7.)

but as the triangle CAB to the triangle BAD, so is the base CA to the base AD, (vi. 1.)

and as the triangle EAD to the triangle DAB, so is the base EA to the base AB; (vI. 1.)

therefore as CA to AD, so is EA to AB: (v. 11.)

wherefore the sides of the triangles ABC, ADE, about the equal angles are reciprocally proportional.

Next, let the sides of the triangles ABC, ADE about the equal angles be reciprocally proportional,

viz. CA to AD as EA to AB.

Then the triangle ABC shall be equal to the triangle ADE.
Join BD as before.

Then because, as CA to AD, so is EA to AB; (hyp.) and as CA to AD, so is the triangle ABC to the triangle BAD: (VI. 1.)

and as EA to AB, so is the triangle EAD to the triangle BAD; (VI. 1.)

therefore as the triangle BAC to the triangle BAD, so is the triangle EAD to the triangle BAD; (v. 11.)

that is, the triangles BAC, EAD have the same ratio to the triangle BAD:

wherefore the triangle ABC is equal to the triangle ADE. (v. 9.) Therefore, equal triangles, &c.

Q.E.D.

PROPOSITION XVI. THEOREM.

If four straight lines be proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means: and conversely, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportionals.

Let the four straight lines AB, CD, E, F be proportionals,
viz. as AB to CD, so E to F.

The rectangle contained by AB, F, shall be equal to the rectangle contained by CD, E.

From the points A, C draw AG, CH at right angles to AB, CD: (1. 11.)

and make AG equal to F, and CH equal to E; (I. 3.)
and complete the parallelograms BG, DH. (1. 31.)

Because, as AB to CD, so is E to F;

and that E is equal to CH, and F to AG;
AB is to CD as CH to AG: (v. 7.)

therefore the sides of the parallelograms BG, DH about the equal angles are reciprocally proportional;

but parallelograms which have their sides about equal angles reciprocally proportional, are equal to one another; (VI. 14.) therefore the parallelogram BG is equal to the parallelogram DH: but the parallelogram BG is contained by the straight lines AB, F; because AG is equal to F;

and the parallelogram DH is contained by CD and E;
because CH is equal to E;

therefore the rectangle contained by the straight lines AB, F, equal to that which is contained by CD and E.

is

And if the rectangle contained by the straight lines AB, F, be equal to that which is contained by CD, E;

these four lines shall be proportional,

viz. AB shall be to CD, as E to F.
The same construction being made,

because the rectangle contained by the straight lines AB, F, is equal to that which is contained by CD, E,

and that the rectangle BG is contained by AB, F;
because AG is equal to F;

and the rectangle DH by CD, E; because CH is equal to E; therefore the parallelogram BG is equal to the parallelogram DH; (ax. 1.)

and they are equiangular:

but the sides about the equal angles of equal parallelograms are reciprocally proportional: (vI. 14.)

wherefore, as AB to CD, so is CH to AG.

But CH is equal to E, and AG to F; therefore as AB is to CD, so is E to F. (v. 7.) Wherefore, if four, &c.

PROPOSITION XVII.

Q.E.D.

THEOREM.

If three straight lines be proportionals, the rectangle contained by the extremes is equal to the square on the mean; and conversely, if the rectangle contained by the extremes be equal to the square on the mean, the three straight lines are proportionals.

Let the three straight lines A, B, C be proportionals,

viz. as A to B, so B to C.

The rectangle contained by 4, C shall be equal to the square on B.

Take D equal to B.

And because as A to B, so B to C, and that B is equal to D;
A is to B, as D to C: (v. 7.)

but if four straight lines be proportionals, the rectangle contained by the extremes is equal to that which is contained by the means; (VI. 16.)

therefore the rectangle contained by A, C is equal to that contained by B, D:

but the rectangle contained by B, D, is the square on B,

because B is equal to D:

therefore the rectangle contained by A, C, is equal to the square on B. And if the rectangle contained by A, C, be equal to the square on B, then A shall be to B, as B to C.

The same construction being made,

because the rectangle contained by A, C'is equal to the square on B, and the square on B is equal to the rectangle contained by B, D, because B is equal to D;

therefore the rectangle contained by A, C, is equal to that contained by B, D:

but if the rectangle contained by the extremes be equal to that contained by the means, the four straight lines are proportionals: (VI. 16.) therefore A is to B, as D to C:

but B is equal to D;

wherefore, as A to B, so B to C.

Therefore, if three straight lines, &c.

Q.E.D.

PROPOSITION XVIII. PROBLEM.

Upon a given straight line to describe a rectilineal figure similar, and similarly situated, to a given rectilineal figure.

Let AB be the given straight line, and CDEF the given rectilineal figure of four sides.

It is required upon the given straight line AB to describe a rectilineal figure similar, and similarly situated, to CDEF.

Join DF, and at the points A, B in the straight line AB, make the angle BAG equal to the angle at C, (1. 23.)

and the angle ABG equal to the angle CDF;

therefore the remaining angle  GB is equal to the remaining angle CFD: (1. 32 and ax. 3.)

therefore the triangle FCD is equiangular to the triangle GAB. Again, at the points G, B, in the straight line GB, make the angle BGH equal to the angle DFE, (1. 23.)

and the angle GBH equal to FDE;

therefore the remaining angle GHB is equal to the remaining angle FED,

and the triangle FDE equiangular to the triangle GBH: then, because the angle AGB is equal to the angle CFD, and BGH to DFE,

the whole angle AGH is equal to the whole angle CFE; (ax. 2.) for the same reason, the angle ABH is equal to the angle CDE: also the angle at A is equal to the angle at C, (constr.)

and the angle GHB to FED:

therefore the rectilineal figure ABHG is equiangular to CDEF: likewise these figures have their sides about the equal angles proportionals;

because the triangles GAB, FCD being equiangular,
BA is to AG, as CD to CF; (VI. 4.)

and because AG is to GB, as CF to FD;

and as GB is to GH, so is FD to FE,

by reason of the equiangular triangles BGH, DFE, therefore, ex æquali, AG is to GH, as CF to FE. (v. 22.) In the same manner it may be proved that AB is to BH, as CD to DE:

and GH is to HB, as FE to ED. (VI. 4.) Wherefore, because the rectilineal figures ABHG, CDEF are equiangular,

and have their sides about the equal angles proportionals,

they are similar to one another. (VI. def. 1.)

Next, let it be required to describe upon a given straight line AB, a rectilineal figure similar, and similarly situated, to the rectilineal figure CDKEF of five sides.

Join DE, and upon the given straight line AB describe the rectilineal figure ABHG similar, and similarly situated, to the quadrilateral figure CDEF, by the former case:

and at the points B, H, in the straight line BH, make the angle HBL equal to the angle EDK,

and the angle BHL equal to the angle DEK;