PROP. XII. PROB.

To find a fourth proportional to three given straight lines (A, B, and c).

A B C

Take two straight lines DE and DF, making any angle EDF, and upon DE make DG equal to A, and GE equal to B; upon DF make DH equal to C; join GH, and from E draw EF parallel to GH and meeting DH produced to F. HF is the required fourth proportional.

For, DG: GE:: DH: HF (vi. Prop. 2); but DG=A, GE=B and DH = C (Const.), and therefore A: B::C: HF (v. Prop. 7), and consequently HF is a fourth proportional to A, B, and C:

G

E

H

F

PROP. XIII. PROB.

To find a mean proportional between two given straight lines (AB, BC).

Let the given lines AB, BC be so placed that they may lie in one continued straight line AC. On AC describe a semicircle ADC, and from B draw BD perpendicular to AC and meeting the circumference. BD is the required mean proportional.

A

B

For, AD and CD being drawn, ADC is a right angle (iii. Prop. 31); and the triangles ABD, DBC, into which the perpendicular divides the triangle ADC, are similar to each other (vi. Prop. 8), their homologous sides being opposite to equal angles; therefore AB: BD:: BD: BC, and consequently BD is a mean proportional between the given lines AB, BC.

PROP. XIV. THEOR.

Equal parallelograms (AB, BC) which have an angle (DBF) of the one equal to an angle (GBE) of the other, have the sides about the equal angles reciprocally proportional (DB: BE:: GB: BF): and parallelograms which have an angle in the one equal to an angle in the other, and the sides about the equal angles reciprocally proportional, are equal.

B

F

E

Let the parallelograms be so placed that the equal angles may be vertically opposite; that is, that DB and BE may lie in the same straight line, and consequently that GB also may lie in the same straight line with BF P (i. Prop. 14). Produce AF and CE till they meet. Then, since the parallelograms AB A and BC are equal, they have the same ratio to the parallelogram FE (v. Prop. 7), or AB: FE:: BC: FE; but the parallelograms AB and FE, having the same altitude, are proportional to their bases (vi. Prop. 1), or AB: FE::DB:BË; and for the same reason, BC: FE :: GB: BF; and therefore DB: BE:: GB: BF (v. Prop. 11), or, the sides about the equal angles are reciprocally proportional.

But if the sides about the equal angles of two parallelograms be reciprocally proportional, then the parallelograms are equal. For, since DB: BE:: GB: BF (Hyp.), it follows that AB:FE:: BC: FE, and the parallelograms AB and BC having the same ratio to the parallelogram FE are therefore equal (v. Prop. 9).

PROP. XV. THEOR.

Equal triangles (ABC, ADE) which have an angle (CAB) of the one equal to an angle (EAD) of the other, have the sides about the equal angles reciprocally proportional: and, triangles which have an angle of the one equal to an angle of the other, and the sides about the equal angles reciprocally proportional, are equal.

B

Let the triangles be so placed that the equal angles may be vertically opposite, that is to say, so that CA and AD may be in the same straight line, and consequently that BA also may lie in the same straight line with AE (i. Prop. 14). Join BD: and then, since the triangles ABC

A

E

and ADE are equal, ABC: ABD :: ADE: ABD (v. Prop. 7) ; but the triangles ABC, ABD having the same altitude, are proportional to their bases, or ABC: ABD :: CA: AD (vi. Prop. 1), and for the same reason, ADE: ABD:: EA: AB, and therefore CA:AD::EA:AB (v. Prop. 11), or, the sides about the angles are reciprocally proportional.

If the sides about the equal angles of two triangles be reciprocally proportional, then the triangles are equal. For, the triangles being placed as before, since CA: AD :: EA : AB, it follows that ABC: ABD :: ADE: ABD, and therefore the triangles ABC and ADE having the same ratio to ABD, are equal (v. Prop. 9).

PROP. XVI. THEOR.

If four straight lines be proportional (AB: CD:: E: F), the rectangle (AB.F) contained by the extremes, is equal to the rectangle (CD.E) contained by the means: and, if the rectangle contained by the extremes be equal to the rectangle contained by the means, the four straight lines are proportional.

G

A

H

B C D

From A and C, the extremities of AB and CD, draw the perpendiculars AG and CH, and make CH E equal to E, and AG equal to F, and complete F the parallelograms BG, DH. Then, since AB:CD::E: F, and CH=E, AG=F, AB: CD::CH: AG, and therefore the parallelograms BG, DH are equal (vi. Prop. 14); but BG is the parallelogram contained by the extremes, since AĞ=F, and DH is the parallelogram contained by the means since CHE; and therefore AB F=CD⚫E. If the rectangle AB·F CD E, then AB CD :: E: F. For, the same construction being made, since AG = F and CH E, the parallelogram BG=DH, and therefore AB: CD :: CH: AG (vi. Prop. 14), that is, AB: CD:: E: F and the four straight lines are proportional.

=

PROP. XVII. THEOR.

If three straight lines be proportional (A: B:: B: C), the rectangle contained by the extremes is equal to the square of the mean: and, if the rectangle contained by the extremes (of three given straight lines) be equal to the square of the mean, the three straight lines are proportional.

A

B

C

D

Since A: B:: B: C, if a fourth line D be assumed equal to B, then A: B:: D: C, and consequently A⚫C, or the rectangle contained by the extremes, is equal to B D or the rectangle contained by the means (vi. Prop. 16). But since D = B, the rectangle B D is the same as B.B, or the square of B; and therefore, if three straight lines are proportional, the rectangle contained by the extremes is equal to the square of the mean.

And conversely, if three straight lines be such that the rectangle contained by the extremes is equal to the square of the mean, these lines are proportional. For, the same construction being made, the sides about the equal parallelograms are reciprocally proportional, or A: B:: D: C (vi. Prop. 14); but DB, and therefore A: B:: B: C, or, the three straight lines are proportional.

PROP. XVIII. PROB.

Upon a given straight line (AB) to describe a rectilineal figure similar to a given rectilineal figure (CEFKD), and similarly situated.

F

K

Resolve the given figure into triangles by drawing the lines DE, DF. At the extremities of the given line AB make ≤ BAG = / DCE, and ZABG=/ CDE: and again, at the extremities of the line GB make BGH=/ DEF, and GBH=/ EDF: in like manner, make < BHL = / DFK, and HBL = / FDK. The figure AGHLB thus constructed is simi- G lar to CEFKD, and similarly situated.

For, the triangles ABG, CDE have BAG

H

=/ DCE, and ▲ ABG = ▲ CDE (Const.), and therefore also ▲ AGB = / CED (i. Prop. 32.); but likewise / BGH=/ DEF (Const.), and if to these be added the equals AGB, CED, then also AGH=/ CEF: and in like manner the remaining angles of the figures may be shown to be equal. And since the triangles ABG, CDE are equiangular, AB: AG :: CD: CE (vi. Prop. 4), and AG : GB::CE: ED; and again, because the triangles BGH, DEF are equiangular, GB: GH :: ED: EF; therefore, ex æquali, AG: GH :: CE: EF (v. Prop. 22). In like manner it may be shown that the remaining sides of the two figures are proportional. Therefore the figure AGHLB constructed on the given line is similar to the given figure, and similarly situated (vi. Def. 1).

PROP. XIX. THEOR.

Similar triangles (ABC, DEF) are to one another in the duplicate ratio of their homologous sides.

E F

B

A

Let A, D be equal angles, and AB, DE homologous sides of the similar triangles ABC, DEF; and on AB, the greater of these lines, take the third proportional AG, so that AB: DE:: DE: AG, and join GC. Then, because the triangles ABC, DEF are similar, AB: AC :: DE: DF (vi. Prop. 4), and by alternation AB: DE:: AC :DF (v. Prop. 16); but AB: DE::DE: AG (Const.), and therefore AC: DF:: DE: AG; consequently, the triangles CAG, DEF, having the angle A in the one equal to the angle D in the other, and the sides about the equal angles reciprocally proportional, are equal (vi. Prop. 15): therefore the triangle ABC: DEF:: ABC: CAG (v. Prop. 7); but ABC: CAG::AB: AG (vi. Prop. 1); and therefore ABC: DEF ::AB: AG, that is to say, the triangles are to one another in the duplicate ratio of their homologous sides AB, DE (v. Def. 11).

COR.-Hence it is manifest that if three straight lines be proportional, as the first is to the third, so is any triangle upon the first, to a similar and similarly described triangle upon the second.