PROPOSITION XIV. THEOREM.

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

1. Let AB, BC be two equal parallelograms, which have the angles at B equal; the sides about the equal angles are reciprocally proportional, that is, DB is to BE as GB to BF.

Let the sides DB, BE A be placed in the same line, then the angles DBG, GBE are equal to two right angles (13.1), there- D fore the angles DBG, DBF are equal to two right an gles, because the angles DBF, GBE are equal; therefore FBG is a straight

F

E

B

C

line. Complete the paral. FE. Because the parals. AB, BC are equal, and FE is another parallelogram,

AB: FE:: BC: FE.

But AB FE::DB: BE (1.6),

and BC

:

FE :: GB: BF;

therefore DB BE:: GB: BF (Propor. 34.)

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

2. Let the sides about the equal angles be reciprocally proportional, DB to BE as GB to BF; the parallelograms AB, BC are equal.

Because DB; BE :: GB : BF,

and DB BE :: AB : FE,

and GB BF :: BC: FE,

AB FE BC: FE (Propor. 34); wherefore the parallelograms AB, BC are equal. Therefore, equal &c. Q. E. D.

Cor. 1. If two triangles, which have one angle of one triangle equal to one angle of the other, be equal to each other, the sides about the equal angles are reciprocally proportional; and conversely. For a triangle is equal to half a paral. of the same base and altitude (41, 1).

ED.

Cor. 2. Equal rectangles have their sides reciprocally proportional; and conversely.

ED.

Cor. 3. Hence if four straight lines be proportional the rectangle contained by the extremes is equal the rectangle contained by the means; and conversely. ED.

Cor. 4. If three straight lines be proportional the rectangle contained by the extremes is equal to the square of the mean; and conversely.

ED.

Cor. 5. Equal triangles, or equal parallelograms, have their bases and altitudes reciprocally proportional; and conversely. ED.

PROPOSITION D. THEOREM.

LEGENDRE.

Two triangles which have one angle of one triangle equal to one angle of the other are to each other as the rectangles contained by the sides about the equal angles.

Let ABC, ADE, be two triangles which have the angles at A equal; the triangle ABC: triangle ADE: AB. AC: AD. AE.

Join BE. The triangles ABE, ADE, whose common vertex is E, have the same altitude, and therefore are to each other as their bases (1. 6), that is,

Let ABC, DEF be two similar triangles, having the angles at A and D equal; and let AC : AB : : DF: DE so that the side AB be homologous to DE (Propor. 22); the triangle ABC : triangle DEF: : ÅB2 : DE'.

From the angles C and F draw the perpendiculars CG and FH; then

by similar triangles ABC, DEF, AB: DE:: AC: DF (4.6), and by similar triangles ACG, DFH, CG: FH:: AC: DF; therefore AB. CG: DE. FH:: ACDF2 (Propor. 42). But the rectangle AB. CG is double of the triangle ABC (41.1), and the rectangle DE. FH is double of the triangle DEF. Consequently the triangle ABC : triangle DEF:: AC: DF2. Therefore, similar triangles &c. Q. E. D.

ED.

Otherwise. The application remaining, the triangle ABC : triangle DEF :: AB. BC: DE. EF (1 Cor. D. 6), and AB: DE :: BC: EF (4. 6). For the ratio of BC: EF in the first proportion substitute the equal ratio of AB: DE, then the triangle ABC triangle DEF AB. AB: DE. DE. fore, similar triangles &c. Q. E. D.

PROPOSITION XX. THEOREM.

There

ED.

Similar polygons may be divided into the same number of similar triangles, having the same ratio to one another which the polygons have; and the polygons have to one another the duplicate ratio of that which their homologous sides have.

Let ABCDE, FGHKL be similar polygons, and let AB be the homologous side to FG; the polygons may be divided into the same number of similar triangles, whereof each has to each the same ratio which the polygons have; and the polygon ABCDE has to FGHKL the duplicate ratio of that which the side AB has to FG.

From any two equal angles E and L draw the lines EB, EC, and LG, LH. Because the polygon ABCDE is similar to FGHKL, the angle BAE is equal to GFL, and BA: AE::GF : FL (Def. 1. 6); therefore the triangle ABE is equiangular (6. 6), and similar to the triangle FGL; wherefore the angle ABE is equal to FGL.

Because the polygons are similar, the whole angles ABC, FGH are equal; therefore the remaining angles EBC, LGH are equal. Now because the triangles ABE, FGL are similar, EB: BA:: LG: GF; and because the polygons are similar, AB: BC:: FG: GH; therefore

EB: BC:: LG: GH (Propor. 42, 43);

therefore the sides about the equal angles EBC, LGH are proportional; therefore the triangles EBC, LGH are equiangular and similar. For the same reason the triangle ECD is similar to LHK. Therefore the similar polygons ABCDE, FGHKL are divided into the same number of similar triangles.

Again, these triangles have, each to each, the same ratio which the polygons have to each other, the antecedents being ABE, EBC, ECD, and the consequents FGL, LGH, LHK; and the polygon ABCDE has to FGHKL the duplicate ratio of that which the side AB has to the homologous side FG.

Because the triangle ABE is similar to FGL, ABE has to FGL the duplicate ratio of that which the side BE has to GL (19. 6). For the same reason the triangle BEC has to GLH the duplicate ratio of that which the side BE has to GL. Therefore the triangle ABE: triangle FGL:: triangle BEC : triangle GLH (Propor. 34).

Again, because the triangle EBC is similar to LGH, EBC has to LGH the duplicate ratio of that which the side EC has to LH. For the same reason the triangle ECD has to LHK the duplicate ratio of that which EC has to LH. Therefore the triangle EBC : triangle LGH :: triangle ECD : LHK (Propor. 34).

But it has been proved that the triangle EBC : triangle LGH :: triangle ABF: triangle FGL. Therefore the triangle ABE : triangle FGL: triangle EBC triangle LGH :: triangle ECD triangle LHK; therefore the triangle ABE: FGL:: polygon ABCDE: FGHKL (Propor. 41).

:

But the triangle ABE: FGL: AB: FG. Therefore the polygon ABCDE: polygon FGHKL :: ABo : FG. Wherefore, similar polygons &c. Q. E. D. !

COR. 1. In like manner it may be proved that similar figures of four sides, or of any number of sides, are to one another in the duplicate ratio of their homologous sides; and it has been proved in triangles. Therefore, universally, similar rectilineal figures are to one another in the duplicate ratio of their homologous sides.

COR. 2. Because all squares are similar figures, the ratio of any two squares to each other is the same with the duplicate ratio of their sides; and hence, also, any two similar rectilineal figures are to each other as the squares of their homologous sides.

A

COR. 3. Two similar triangles, or two similar polygons, are to each other as any rectilineal figure described on any side of one is to a similar figure similarly described on the homologous side of the other.

For the two given figures, and the two similar figures thus similarly described, will have to each other the same duplicate ratio of that which their homologous sides have to each other. ED.

PROPOSITION XXIII. THEOREM.

Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides about the equal angles; "that is, they are to one another as the rectangles contained by the sides about the equal angles."

Let AC, CF be two equiangular parallelograms, having the angle BCD equal to ECG; the ratio of the paral. AC to the