angles proportionals. If not equal, let them stand as indicated. Draw DG parallel to AB.

Because the external angle AED is equal to the internal opposite angle ABC, therefore the straight lines ED and BC are parallel (P. 8), therefore BE is to EA as CD is to DA (P. 17), therefore also, by composition, BE and EA together are to EA, as CD and DA together are to DA (16, cor. 2), that is, BA to EA as CA to DA. Therefore alternately BA to CA as EA to DA (16, cor. 1).

Because DG is parallel to AB. Therefore CD to DA as CG to GB (P. 17). And, by composition, CD, DA together are to DA; as CG, GB together are to GB (16, cor. 2), that is, CA to DA as BC to GB. But GB is equal to ED (P. 11); therefore CA to DA as BC to ED. And alternately CA to BC as DA to DE (16, cor 1).

Because it has been proved that

And also

BA to CA as EA to DA

CA to BC as DA to DE

Therefore (16, cor. 6) BA to BC as EA to DE. ·
Wherefore the sides are proportionals.

Conversely, let the triangles ABC and AED have their sides proportionals, so that AB to BC as AE to ED and AB to AC as AE to AD and consequently AC to BC as AD to DE. The angle ABC is equal to the angle AED, the angle BAC to the angle EAD and the anglə ACB to the angle ADE.

Place the trianglo AED upon the side CA, so that the point D may be on C and the side DA on the side CA Draw AG parallel to AB. Join EG.

It can be shown that the triangle ABC and AGC are equiangular and therefore have their sides about the

equal angles proportionals, i.e., AB to BC as AG to GC, AB to AC as AG to AC and also AC to BC as AC to CG. Because AB to AC as AG to AC, and AB to AC as AE to AC (hyp.) and because ratios which are equal to the same ratio, are equal to one another, therefore AG to AC as AE to AC. Because magnitudes having the same ratio to a magnitude of the same kind are equal, therefore AG and AE are equal. In the same manner it can be proved that CG and EC are equal.

Because AG and AE are equal, therefore the angles AGE and AEG are equal (P. 4). For the same reason the angles CGE and CEG are equal. Therefore the whole angles AGC and AEC are equal (Ax. 2). Wherefore the triangles AGC and AEC are equiangular (P. 3) : viz., the angle GAC is equal to the angle CAE, the angle AGC to the angle AEC and the angle ACG to the angle ACE. But it has been shown that the angles GAC, AGC and ACG are respectively equal to the angles BAC, ABC and ACB, wherefore the angles CAE, AEC and ACE are respectively equal to them (Ax. 1). Wherefore the

sides, &c. Q. E. D.

Cor. Equal sided triangles are equiangular, and equal (E. 1. 8).

PROP. XIX. THEOREM. (E. 6. 14, 15).

Equal triangles and parallelograms having an angle of the one, équal to an angle of the other, have their sides about the equal angles, reciprocally proportional. And conversely triangles and parallelograms having an angle of the one equal to an angle of the other and their sides about the equal angles reciprocally proportional, are equal to one another.

First, let ADC and EDB be two triangles, having their

A

angles ADC and BDE equal. The sides DC, DA and BD, DE are alternately proportional, i.e., DC to DB as DE to DA. Place the triangle EDB so that the base BD may coincide with a part produced of CD, ie., lie in the same straight line with CD. Let the angle BDE stand opposite to the angle ADC. Join AB.

Because the triangles ADC and EDB are equal, therefore they have the same ratio to the triangle ABD (16, cor. 6).

B

E

D

But ADC to ADB as DC to BD (P. 16). And EDB to ADB as ED to DA. Therefore CD to DB as ED to DA

(P. 16, cor. 4).

Conversely, if CD to DB as ED to DA, then because CD to BD as ADC to ABD, and ED to DA as EBD to ABD (P. 16). Therefore ADC to ADB as EDB to ADB (16, cor. 4). Therefore ADC and EDB are equal (16, cor. 6). Next, let AC and FD be two equal parallelograms, having the angles at B and E equal. AB to DE as EF to BC. Join AB, DF.

Because the triangles ABC and DEF are halves of the equal things AC and DF (P. 13), therefore they are equal (Ax: 5). But they have their angles at B and E equal. Therefore according to the first case AB to DE as EF to BC. That is, the parallelograms AC, DF have their sides AB, BC and DE, EF reciprocally i.e., alternately, proportional.

Conversely, if AB is to DE as EF is to BC and the angle at B and E equal, then according to the first case the triangles ABC and DEF are equal. And

because AC and DF are their doubles (P. 13), therefore are equal (Ax. 4).

PROP. XX. THEOREM. (E. 1. 43, 6. 24).

Parallelograms about the diagonal of any parallelogram, are similar to the whole, and to one another and the complements about the diognal are egual.

E

A H

K

D

Let BD be a parallelogram, of which the diagonal is AC; and FG and EH parallelograms about the diagonal. Parallelograms FG and EH are similar to the whole parallelogram BD, and to each other. KB and KD are other parallelograms which make up the B whole figure BD, and are therefore called the complements These complements are equal.

G

First because BC and EK are parallels, the angle ABC is equal to the angle AEK (P. 8). Because DC and HK are parallels, the angle ADC is equal to the angle AHK (P. 8). Because each of the angles BCD and HKE is equal to the opposite angle BAD (P. 11). Therefore they are equal to one another. Wherefore the parallelograms BD and HE are equiangular. Because the angle ADC is equal to the angle AHK and the angle DAC common to the two triangles DAC and AHK.. They are equiangular to one another. Therefore AD is to DC, as AH is to HK (P. 18). Because the opposite sides of parallelograms are equal to one another (P. 11). Therefore AD is to AB, as AH is to AE (16. cor. 6); BC to CD, as EK to HK; and CB to BA as EK to EA. Therefore the sides of the parallelograms BD and EH about the equal angles are propotionals; and they are similar to one another.

For the same reason the parallelogram DB is similar to the parallelogram HE. Because each of the parallelograms EH and FG is similar to the parallelogram BD, that is, has its angles equal to the angles of BD and its sides and those of BD about the equal angles proportionals, therefore EH and FG have their angles equal (Ax. 1) and the sides about those angles proportionals (16, cor. 4), wherefore parallelograms EH, FG and BD are similar.

Next, because FG and HK are similar, therefore FK is to KH as FK is to GK (hyp.) and also the angle EKG is equal to the angle HKF (P. 2), therefore the parallelograms EG and KD are equal (P. 19), wherefore parallelograms &c. Q. E. D.

M

To describe a rectangle equal to a given rectilineal figure. Let ABCD be a rectilineal figure. It is required to make a retangle equal to it. Join BD.

B F

Bisect BC at F and draw FD perpendicular from the point of section. First, let FD and BD be terminated in the same point D. Draw CE parallel to FD and DE parallel to FC. It may be proved that FE is a rectangle.

Because the triangles DBF and DFC are upon equal bases and between the same parallels, therefore they are equal (P. 15) and DBC is double of DFC. Because FE is a parallelogram and DFC a triangle upon the same base and between the same parallels, therefore the parallelogram is double of DFC (P. 13). Therefore the paral lelogram FE is equal to the triangle DBC (Ax. 4).