And DF is common to the two triangles EDF, GDF; therefore the two sides ED, DF are equal to the two sides GD, DF, each to each; and the angle EDF (Constr.) is equal to the angle GDF; wherefore the base EF is equal to the base FG (4. 1.), and the triangle EDF to the triangle GDF, and the remaining angles to the remaining angles, each to each, which are subtended by the equal sides; therefore the angle DFG is equal to the angle DFE, and the angle at G to the angle at E:

But the angle DFG is equal to the angle ACB; therefore the angle ACB is equal to the angle DFE: and the angle BAC is equal to the angle EDF (Hyp.); wherefore also the remaining angle at B is equal to the remaining angle at E: therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, etc.

Q. E. D.

PROPOSITION VII.

THEOR. If two triangles have one angle of the one equal to one angle of the other, and the sides about two other angles proportionals; then, if each of the remaining angles be either less, or not less, than a right angle, or if one of them be a right angle, the triangles shall be equi angular, and have those angles equal about which the sides are pro portionals.

Let the two triangles ABC, DEF have one angle in the one equal to one angle in the other, viz., the angle BAC to the angle EDF, and the sides about two other angles ABC, DEF proportionals, so that AB is to BC, as DE to EF; and, in the first case, let each of the remaining angles at C, F be less than a right angle: the triangle ABC is equiangular to the triangle DEF, viz., the angle ABC is equal to the angle DEF, and the remaining angle at C equal to the remaining angle at F. For if the angles ABC, DEF be not equal, one of them is greater than the other: let ABC be the greater, and at the point B, in the straight line AB, make the angle ABG equal

to the angle (23. 1.) DEF:

C E

F

And because the angle at A is equal to the angle at D, and the angle ABG to the angle DEF, the remaining angle AGB is equal (32. 1.) to the remaining angle DFE: therefore the triangle ABG is equiangular to the triangle DEF; wherefore (4. vi.) as AB is to BG, so is DE to EF: but as DE to EF, so, by hypothesis, is AB to BC; therefore as AB to BC, so is AB to BG (11. v.): and because AB has the same ratio to each of the lines BC, BG; BC is equal (9. v.) to BG; and therefore the angle BGC is equal to the angle BCG (5. 1.):

But the angle BCG is, by hypothesis, less than a right angle; therefore also the angle BGC is less than a right angle, and the adjacent angle AGB must be greater than a right angle (13. 1.):

But it was proved that the angle AGB is equal to the angle at F; therefore the angle at F is greater than a right angle: but, by the hypothesis, it is less than a right angle; which is absurd. Therefore the angles ABC, DEF are not unequal, that is, they are equal: and the angle at A is equal (Hyp.) to the angle at D; wherefore the re

maining angle at C is equal to the remaining angle at F: therefore the triangle ABC is equiangular to the triangle DEF.

A

Next, let each of the angles at C, F be not less than a right angle; the triangle ABC is also in this case equiangular to the triangle DEF. The same construction being made, it may be proved in like manner that BC is equal to BG, and the angle at C equal to the angle BGC: but the angle at C is not less than a right angle; therefore the angle BGC is not less than a right angle: wherefore two angles

3

of the triangle BGC are together not less than two right angles, which is impossible (17. 1.); and therefore the triangle ABC may be proved to be equiangular to the triangle DEF, as in the first case. Lastly, let one of the angles at C, F, viz., the angle at C, be a right angle: in this case likewise the triangle ABC is equiangular to the triangle DEF.

For if they be not equiangular, make, at B the point B of the straight line AB, the angle ABG equal to the angle DEF; then it may be proved, as in the first case, that BG is equal to BC:

But the angle BCG is a right angle, therefore (5. 1.) the angle BGC is also a

B

A

F

right angle; whence two of the angles of the triangle BGC are together not less than two right angles, which is impossible (17. 1.): therefore the triangle ABC is equiangular to the triangle DEF. Wherefore, if two triangles, etc. Q. E. D.

PROPOSITION VIII.

THEOR. In a right-angled triangle, if a perpendicular be drawn from the right angle to the base, the triangles on each side of it are similar to the whole triangle, and to one another.

Let ABC be a right-angled triangle, having the right angle BAC; and from the point A, let AD be drawn perpendicular to the base BC: the triangles ABD, ADC are similar to the whole triangle ABC, and to one another.

A

Because the angle BAC is equal to the angle ADB (11 Ax.) each of them being a right angle, and that the angle at B is common to the two triangles ABC, ABD; the remaining angle ACB is equal to the remaining angle BAD (32. 1.): therefore the triangle ABC is equiangular to the triangle ABD, and the sides about their equal

ח

angles are proportionals (4. vI.); wherefore the triangles are similar (1 Def. VI.):

In the like manner it may be demonstated, that the triangle ADC is equiangular and similar to the triangle ABC:

And the triangles ABD, ACD, being both equiangular and similar to ABC, are equiangular and similar to each other. Therefore, in a right-angled, etc. Q. E. D.

COR. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the base, is a mean proportional between the segments of the base, and also that each of the sides is a mean proportional between the base, and its segment adjacent to that side: because in the triangles BDA, ADC, BD is to DA, as DA to DC (4. vI.); and in the triangles ABC, DBA, BC is to BA, as BA to BD (4. vI.); and in the triangles ABC, ACD, BC is to CA, as CA to CD (4. vr.).

PROPOSITION IX.

PROB. From a given straight line to cut off any part required. Let AB be the given straight line; it is required to cut off any part from it.

From the point A draw a straight line AC, making any angle with AB; and in AC take any point D, and take AC the same multiple of AD, that AB is of the part which is to be cut off from it: join BC, and draw DE parallel to it: then AE is the part required to be cut off.

Because ED is parallel to one of the sides of the triangle ABC, viz. to BC; as CD is to DA, so is (2. v1.) BE to EA; and by composition (18. v.), CA is to AD, as BA to AE:

B

D

C

But CA is a multiple of AD; therefore (D. v.) BA is the same multiple of AE; whatever part, therefore, AD is of AC, AE is the same part of AB: wherefore, from the straight line AB, the part required is cut off. Which was to be done.

PROPOSITION X.

PROB. To divide a given straight line similarly to a given divided straight line, that is, into parts that shall have the same ratios to one another which the parts of the divided given straight line have.

Let AB be the straight line given to be divided, and AC the divided line; it is required to divide AB similarly to AC.

Let AC be divided in the points D, E; and let AB, AC be placed so as to contain any angle, and join BC; and through the points D, E draw (31. 1.) DF, EG parallels to it; and through D draw DHK parallel to AB:

Therefore each of the figures FH, HB is a parallelogram: wherefore DH is equal (34. 1.) to FG, G

and HK to GB:

F

B

H

D

And because IIE is parallel to KC, one of the sides of the triangle DKC, as CE to ED, so is (2. vI.) KH to HD: but KH is equal to BG, and HD to GF; therefore, as CE to ED (7. v.), so is BG to GF:

Again, because FD is parallel to GE, one of the sides of the triangle AGE, as ED to DA, so is GF (2. vi.) to FA: but it has been proved that CE is to ED, as BG to GF; and as ED to DA, so is GF to FA. Therefore, the given straight line AB is divided similarly to AC. Which was to be done.

PROPOSITION XI.

PROB. To find a third proportional to two given straight lines.

Let AB, AC be the two given straight lines, and let them be placed so as to contain any angle; it is required to find a third proportional to AB, AC.

Produce AB, AC, to the points D, E; and make BD equal to AC (3. 1.); and having joined BC, through D, draw DE parallel to it (31. 1.):

Because BC is parallel to DE, a side of the triangle ADE, AB is (2. vi.) to BD, as AC to CE:

But BD is equal to AC; as therefore AB to (7. v.) AC, so is AC to CE. Wherefore, to the two given straight lines AB, AC, a third proportional CE is found. Which was to be done.

PROPOSITION XII.

B

C

PROB. To find a fourth proportional to three given straight lines.

Let A, B, C be three given straight lines; it is required to find a fourth proportional to A, B, C.

Take two straight lines, DE, DF, containing any angle EDF; and upon these make DG equal to A, GE equal

to B, and DH equal to C; and having joined GH, draw EF parallel (31. 1.) to it through the point E:

C

A

B

C

H

And because GH is parallel to EF, one of the sides of the triangle DEF, DG is to GE, as DH to HF (2. vi.): but DG is equal to A, GE to B, and DH to C; therefore, as A is to B (7. v.), so is C to HF. Wherefore, to the three given straight lines A, B, C, a fourth proportional HF is found. Which was to be done.

E

PROPOSITION XIII.

PROB. To find a mean proportional between two given straight

lines.

Let AB, BC be the two given straight lines; it is required to find a mean proportional between them.

Place AB, BC in a straight line, and upon AC describe the semicircle ADC; and from the point B draw (11. 1.) BD at right angles to AC, and join AD, DC.

Because the angle ADC in a semicircle is a A right angle (31. III.), and because in the right

angled triangle ADC, DB is drawn from the right angle perpendicular to the base, DB is a mean proportional between AB, BC the segments of the base (Cor. 8. vI.). Therefore, between the two given straight lines AB, BC, a mean proportional DB is found. Which was to be done.

PROPOSITION XIV.

THEOR. 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 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, and let the sides DB, BE be placed in the same straight line; wherefore also FB, BG are in one straight line (14. 1.). The sides of the parallelograms AB, BC about the equal angles are reciprocally proportional; that is, DB is to BE, as GB to BF.

Complete the parallelogram FE:

And because the parallelogram AB is equal to BC (Hyp.), and that FE is another parallelogram, AB is to FE, as BC to FE (7. v.):

But as AB to FE, so is the base DB to BE (1. vI.); and as BC to FE, so is the base of GB to BF; therefore, as DB to BE, so is GB to BF (11. v.). Wherefore, the sides of the parallelograms AB, BC about their equal angles are reciprocally proportional.

But, let the sides about the equal angles be reci- D procally proportional, viz. as DB to BE, so GB to BF; the parallelogram AB is equal to the parallelogram BC.

F

E

B

Because, as DB to BE, so is GB to BF; and as DB to BE (1. vi.), so is the parallelogram AB to the parallelogram FE; and as GB to BF, so is the parallelogram BC to the parallelogram FE; therefore, (11. v.) as AB to FE, so BC to FE: wherefore the parallelogram AB is equal (9. v.) to the parallelogram BC. Therefore, equal parallelograms, etc. Q. E. D.

PROPOSITION XV.

THEOR. 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 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; the sides about the equal angles of the triangles are reciprocally proportional; that is, CA is to AD, as EA to АВ.

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 (14. 1.); and join BD.

B

D

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 triangle EAD to triangle DAB (7. v.):

But as triangle CAB to triangle BAD, so is the base CA to AD (1. vr.); and as triangle EAD to triangle DAB, so is the base EA to AB (1. vI.); as therefore CA to AD, so is EA to AB (11. v.):