Because C A is to AD, as E A is to A B (Hyp.). But CA is to AD, as the triangle ABC is to the triangle BAD (VI. 1); and EA is to A B, as the triangle EAD is to the triangle B A.D. Therefore the triangle BAC is to the triangle B AD, as the triangle E A D is to the triangle BAD (V. 11). Wherefore the triangle ABC is equal to the triangle ADE (V. 9). Therefore, equal triangles, &c. Q. E. D.

Corollary.-Equal triangles which have an angle in the one supplementary to an angle in the other, have their sides about the supplementary angles reciprocally proportional; and conversely, triangles which have an angle in the one supplementary to an angle in the other, and their sides about the supplementary angles reciprocally proportional, are equal to one another.

PROP. XVI. THEOREM.

I 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.

First, let the four straight lines AB, CD, E and F be proportionals, viz., AB to CD, as E to F. The rectangle contained by A B and F, is equal to the rectangle contained by CD and E.

F

From the points A and C draw AG and CH at E. right angles to AB and CD (I. 11). Make AG equal to F, and CH equal to E (I. 3). Complete the parallelograms BG and DH (I. 31).

G

A

B C

D

Because AB is to CD, as E is to F, and E is equal to CH, and F to AG. Therefore AB is to CD as CH is to AG (V. 7); and the sides of the parallelograms BG and D H about the equal angles are reciprocally proportional. Therefore the parallelogram BG is equal to the parallelogram DH (VI. 14). But the parallelogram BG is contained by the straight lines A B and F; because A G is equal to F. And the parallelogram DH is contained by the straight lines CD and E; because CH is equal to E. Therefore the rectangle contained by the straight lines A B and F, is equal to the rectangle contained by the straight lines CD and E.

Next, if the rectangle contained by the straight lines A B and F, be equal to the rectangle contained by the straight lines CD and E; the four straight lines are proportionals, that is, AB is to CD, as E is to F.

The same construction being made, the rectangle contained by the straight lines AB and F, is equal to the rectangle contained by CD and E (Hyp.). But the rectangle B G is contained by AB and F; because AG is equal to F. And the rectangle DH is contained by the straight lines CD and E; because CH is equal to E. Therefore the parallelogram BG is equal to the parallelogram D H (I. Ax. 1); and they are equiangular. But the sides about the equal angles of equal parallelograms are reciprocally proportional (VI. 14). Therefore, A B is to CD, as CH is to AG. But CH is equal to E, and AG to F. Therefore AB is to CD, as E is to F (V. 7). Wherefore if four, &c. Q. E. D.

Corollary.-Equal triangles and parallelograms have their bases and altitudes reciprocally proportional; and conversely, triangles and parallelograms which have their bases and altitudes reciprocally proportional, are equal.

PROP. XVII. THEOREM.

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

First, let the three straight lines A, B, and C be proportionals, viz., A to B, as B to C. The rectangle contained by A and C is equal to the square of B.

Take the straight line D equal to the straight line B.

A

B

D

C

A

C

D

B

Because A is to B, as B is to C, and B is equal to D. Therefore A is to B, as D is to C (V. 7). But when four straight lines are proportionals, the rectangle contained by the extremes is equal to the rectangle contained by the means (VI. 16). Therefore the rectangle contained by A and C is equal to the rectangle contained by B and D. But the rectangle contained by B and D, is the square of B, because B is is equal to D. Therefore the rectangle contained by A and C, is equal to the square of B.

Next, if the rectangle contained by A and C, be equal to the square of B, the three straight lines are proportionals, that is, A is to B, as B is to C.

The same construction being made, the rectangle contained by A and C is equal to the square of B (Hyp.). But the square of B is equal to the rectangle contained by B and D, because B is equal to D (Const.). Therefore the rectangle contained by A and C, is equal to that contained by B and D (I. Ax. 1). But if the rectangle contained by the extremes bc equal to that contained by the means, the four straight lines are proportionals (VI. 16). Therefore A is to B, as D is to C. But B is equal to D. Therefore A is to B, as B is to C. Wherefore, if three straight lines, &c. Q. E. D.

Exercise 1.-To construct a square equal to a parallelogram, whose base and altitude are given.

Exercise 2.-To construct a square equal to a triangle, whose base and altitude are given.

Exercise 3.-Demonstrate the 47th proposition of the first Book, in the manner suggested in the annotation to Prop. VIII.

DEFINITION. Similar figures are said to be similarly situated, when their homologous sides are parallel and drawn in the same direction. This definition is given to explain the enunciation of the next proposition.

PROP. XVIII. PROBLEM.

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

First let AB be the given straight line, and CDEF the given rectilineal figure of four sides. It is required to describe upon the given

straight line AB a rectilineal figure similar, and similarly situated, to the rectilineal figure CDEF.

H

T

K

B C D

Join DF, and at the points A and B in the straight line AB, make the angle BAG equal to the angle DCF (I. 23), and the angle ABG equal to the angle CDF. Because the remaining angle AGB is equal to the remaining angle CFD (I. 32, and Ax. 3). A Therefore the triangle G AB is equiangular to the triangle FCD. Again, at the points G and B, in the straight line G B, make the angle BGH equal to the angle DFE (I. 23), and the angle GBH equal to the angle FDE. Because the remaining angle GHB is equal to the remaining angle FED. Therefore the triangle GBH is equiangular to the triangle FDE. Because the angle AGB is equal to the angle CFD, and the angle BGH to the angle DFE. Therefore the whole angle AGH is equal to the whole angle CFE (I. Ax. 2). For the same reason, the whole angle ABH is equal to the whole angle CDE. Also the angle BAG is equal to the angle DCF (Const.), and the angle GHB to the angle FED. Therefore the rectilineal figure ABHG is equiangular to the rectilineal figure CDEF. Likewise these figures have their sides about the equal angles proportionals. Because the triangles GAB and FCD are equiangular, BA is to AG, as DC is to CF (VI. 4), and AG is to GB, as CF is to FD. Because the triangles BGH and D FE are equiangular, therefore GB is to GH, as FD is to FE. Wherefore, ex æquali, AG is to G H as CF is to FE (V. 22). In the same manner, it may be proved that A B is to B H, as CD is to DE and GH is to HB, as FE is to ED (VI. 4). Because the rectilineal figures A BHG and CDEF are equiangular, and have their sides about the equal angles proportionals. Therefore they are similar to one another (VI. Def. 1).

Next, let it be required to describe upon the given straight line AB, a rectilineal figure similar, and similarly situated, to the rectilinea. 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. At the points B and H, in the straight line B H, make the angle HBL equal to the angle E DK, and the angle BHL equal to the angle DEK (I. 23).

Because the two angles HBL and B HL are equal to the two angles EDK and DE K. Therefore the remaining angle BL H is equal to the remaining angle DKE (I. 32). Because the figures ABHG and CDEF are similar, the angle GHB is equal to the angle FED (VI. Def. 1), but the angle B HL is equal to the angle DEK. Therefore the whole angle GHL is equal to the whole angle FEK. For the same reason the whole angle AB L is equal to the whole angle CDK. Wherefore the five-sided figures AGHLB and CFEKD are equiangular. Because the figures AGHB and CFED are similar. Therefore G H is to HB, as FE is to ED (VI. Def. 1). But HB is to HL as ED is to EK (VI. 4). Therefore, ex æquali, G H is to HL as FE is to EK (V. 22). For the same reason, AB is to BL, as CD is to DK. Because the triangles B L H and D K E are equiangular. Therefore BL is to LH, as

DK is to KE (VI. 4). Because the five-sided figures A GHLB and CFEKD are equiangular, and have their sides about the equal angles proportionals. Therefore they are similar to one another. In the same manner, a rectilineal figure may be described upon a given straight line similar to a given rectilineal figure of six sides; and so on. Q. E. F.

This problem is more easily effected in practice, by placing the given straight line AB in the same straight line with, or parallel to, the assumed base CD, and drawing straight lines parallel to the sides of the figure, and of the triangles into which it is divided by the straight lines drawn from the point D to its different angles. The proof is thus also more easily established.

PROP. XIX. THEOREM.

Similar triangles are to one another in the duplicate ratio of their homologous sides.

Let A B C and DEF be similar triangles, and let the angle ABC be equal to the angle DEF, and let A B be to B C, as DE to EF, so that the side B C is homologous to EF (V. Def. 12). The triangle ABC has to the triangle D E F, the duplicate ratio of that which the side BC has to the side E F.

Take B G a third proportional to BC and EF (VI. 11); so that B C is to EF, as E F to BG. Join GA.

Because AB is to BC, as D E is to EF (Hyp.). Therefore, alternately, AB is to DE,

A

CE

as BC is to EF (V. 16). But B C is to EF, B G as EF is to BG (Const.). Therefore A B is to DE, as EF is to BG (V. 11). Wherefore the sides of the two triangles ABG and DE F, which are about the equal angles ABG and DEF, are reciprocally proportional. But triangles, which have the sides about two equal angles reciprocally proportional, are equal to one another (VI. 15). Therefore the triangle ABG is equal to the triangle DEF. Because B C is to EF, as EF is to BG; and if three straight lines be proportionals, the first is said to have to the third, the duplicate ratio of that which it has to the second (V. Def. 10), __ Therefore BC has to B G the duplicate ratio of that which BC has to EF. But BC is to B G, as the triangle ABC is to the triangle ABG (VI. 1). Therefore the triangle ABC has to the triangle ABG, the duplicate ratio of that which BC has to E F. But the triangle ABG has been proved equal to the triangle DEF. Therefore also the triangle ABC has to the triangle DEF, the duplicate ratio of that which the side BC has to the side È F. Therefore similar triangles, &c. Q. E. D.

COROLLARY.-From this it is manifest, that if three straight lines be proportionals, as the first is to the third, so is any triangle upon the first to a similar and similarly described triangle upon the second.

This proposition is one of the most important in the Elements, and as such ought to be carefully studied and remembered by the student. It is the basis on which the comparison of the areas of similar rectilinear figures is founded : and it thus furnishes the rule by which, when the area of one rectilineal figure or polygon is known, the areas of all similar rectilineal figures or polygons are obtained.

Similar polygons are divisible into the same number of similar triangles; these triangles have to one another the same ratio that the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous sides have.

Let ABCDE and FGHKL be similar polygons, and let AB be the side homologous to FG. The polygons ABCDE and FGHKL are divisible into the same number of similar triangles. These triangles have to one another the same ratio which_the_polygons have. And the polygon ABCDE has to the polygon FGHKL the duplicate ratio of that which the side A B has to the side F G. Join BE, EC, GL, and LH.

D

A

M

L

K H

G

Because the polygon ABCDE is similar to the polygon FGH KL, the angle BAE is equal to the angle GFL, and BA is to AE, as GF is to FL (VI. Def. 1). Because the triangles ABE and FGL have an angle in the one, equal to an angle in the other, and their sides about the equal angles proportionals. Therefore the triangle ABE is equiangular and similar to the triangle FGL (VI. 6 and 4). Wherefore the angle ABE is equal to the angle FGL. Because the polygons are similar, the whole angle ABC is equal to the whole angle FGH (VI. Def. 1). Therefore the remaining angle EBC is equal to the remaining angle LGH (I. 32, and Ax. 3). Because the triangles ABE and FGL are similar, E B is to BA, as LG is to GF (VI. 4). Because the polygons are similar, A B is to BC, as FG is to GH (VI. Def. 1). Therefore, ex æquali, E B is to BC, as LG is to GH (V. 22). Because the sides about the equal angles EBC and LGH are proportionals. Therefore the triangle EBC is equiangular, and similar to the triangle L G H (VI. 6 and 4). For the same reason, the triangle E CD is similar to the triangle L HK. Therefore the similar polygons A B C D E and F G H K L are divided into the same number of similar triangles.

Again, because the triangles are similar, the triangle ABE has to the triangle F G L the duplicate ratio of that which the side B E has to the side GL (VI. 19). For the same reason, the triangle BEC has to the triangle G L H the duplicate ratio of that which BE has to G L. Therefore the triangle AB E is to the triangle F G L, as the triangle BEC is to the triangle GLH (V. 11). Because the triangles are similar, the triangle EBC has to the triangle LGH the duplicate ratio of that which the side E C has to the side L H. For the same reason, the triangle ECD has to the triangle LHK the duplicate ratio of that which EC has to LH. Therefore the triangle E B C is to the triangle LGH, as the triangle ECD is to the triangle LHK (V. 11). But it has been proved, that the triangle EBC is to the triangle LGH, as the triangle ABE is to the triangle F G L. Therefore the triangle ABE is to the triangle F GL, as the triangle E B C is to the triangle LG H, and as the triangle ECD is to the triangle LHK. But one of the antecedents is to one of the consequents, as all the antecedents are to all the consequents (V. 12). Therefore the triangle ABE is to the triangle F G L, as the polygon ABCDE is to the polygon FGHK L. But the triangle