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CAD : and because the straight line FAE meets the parallels AD, FC, the angle CFA is equal to DAE: therefore also ACF, CFA are (I. ax. 1.) equal to one another, because they are respectively equal to the equal angles DAC, DAE; and consequently (I. 6.) AF is equal to AC. Then (VI. 2.) because AD is parallel to FC, a side of the triangle BCF, BD: DC :: BA: AF: but AF is equal to AC; as therefore BD:DC::BA:AC.

2. Let now BD : DC :: BA: AC, and join AD; the angle CAD is equal to DAE.

The same construction being made, because BD:DC::BA: AC; and that (VI. 2.) BD:DC::BA: AF; therefore (V. 11.) BA: AC::BA : AF; wherefore (V. 9.) AC is equal to AF, and (I. 5.) the angle AFC to ACF. But (I. 29.) the angle AFC is equal to EAD, and ACF to CAD; therefore also (1.ax. 1.) EAD is equal to CAD. Wherefore, if an exterior angle, &c.

Cor. If G be the point in which BC is cut by the straight line bisecting the angle BAC,

we have (VI. 3.) BG:GC::BA: AC;

and by this proposition, BD: DC:: BA: AC; whence, (V. 11.) BG: GC:: BD: DC, and therefore (VI. def. 5.) BD is divided harmonically in G and C.

PROP. IV. THEOR. * The sides about the equal angles of equiangular triangles are proportionals; and those which are opposite to the equal angles are homologous sides, that is, are the antecedents or consequents of the ratios.

Let ABC, DCE be equiangular triangles, having the angle ABC equal to DCE, and ACB to DEC, and consequently (1. 32. cor. 5.) BAC equal to CDE: the sides about the equal angles are proportionals; and those are the homologous sides which are opposite to the equal angles.

Let the triangles be placed on the same side of a straight line BE, so that sides BC, CE, which are opposite to equal angles, may be in that straight line and contiguous to one another; and so that the equal a angles ABC, DCE at the extremities of those sides may not be adjacent. Then, because (I. 17.) the angles ABC, ACB are together less than two right angles, ABC and DEC, which (hyp.) is equal to ACB, are also less than two right angles ; wherefore (1. ax. 12.) BA, ED produced will meet: let them be produced and meet in F. Again,

* This important proposition is a generalization of the 26th of the first book, as the fifth and sixth of this book are of the 8th and 4th of the first.

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(I. 28.) because the angle ABC is equal to DCE, BF is parallel
to CD; and, because the angle ACB is equal to DEC, AC is
parallel to FE. Therefore FACD is (I. def. 24.)
a parallelogram; and consequently (1. 34.) AF
is equal to CD, and AC to FD. Now, (VI. 2.) A
because AC is parallel to FE, one of the sides of
the triangle FBE,

BA: AF:: BC: CE.
But AF is equal to CD; therefore (V. 7.)

as BA : CD::BC: CE; and alternately (V. 16.) as AB: BC :: DC : CE. Again (VI. 2.) because CD is parallel to BF, as BC: CE:: FD : DE: but FD is equal to AC; therefore, as BC: CE :: AC : DE; and, alternately,

as BC: CA:: CE: ED. Therefore, because it has been proved that

AB: BC :: DC: CE, and as BC:CA :: CE: ED ;' exæquo, (V. 22.) BA: AC::CD: DE. Therefore, the sides, &c.

OTHERWISE. Let the triangles be ABC and ADE; and, the angles BAC, DAE being equal, let the triangles be placed so that the sides AB, AD, which are opposite to equal angles C and E, may be in the same straight line, and contiguous, and that the triangles may be on opposite sides of the straight line BAD: then (I. 14. cor.) AC, AE are in the same straight line. Again, since (hyp.) the angles C and E are equal, the straight lines BC, ED are (1. 27.) parallel; and (VI. 2. cor. part. 1.) BA : AC :: DA: AE. In like manner, by varying the positions of the triangles, it would be shown that the sides about B and D, and those about C and E are proportional.

Schol. Hence (VI. def. 1.) equiangular triangles are similar.

Cor. If two angles of one triangle be respectively equal to two angles of another, their sides are proportional, and the sides opposite to equal angles are homologous. For (1. 32. cor. 5.) the remaining angles are equal, and therefore the triangles are equiangular.

PROP. V. THEOR. If the sides of two triangles, about each of their angles, be proportionals, the triangles are equiangular, and have their equal angles opposite to the homologous sides.

Let the triangles ABC, DEF, have their sides proportionals, so that AB:BC :: DE: EF; and BC:CA:: EF:FD; and consequently, ex æquo, BA: AC:: ED: DF; the triangles are equi

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angular, and the equal angles are opposite to the homologous sides, viz., the angle ABC equal to DEF, BCA to EFD, and BAC to EDF.

At the points E, F, in the straight line EF, make (I. 23.) the angle FEG equal to B, and EFG equal to C. Then (VI. 4. cor.) the triangles ABC, GEF have their sides opposite to the equal angles proportionals ; wherefore

AB : BC :: GE: EF; but (hyp.)

AB: BC ::DE: EF. Therefore (V. 11.) DE: EF:: GE: EF; whence, since DEand GE have the same ratio to EF, they are (V. 9.) equal. It may be shown in a similar manner that DF is equal to FG: and because, in the triangles DEF, GEF, DE is equal to EG, EF common, and DF equal to GF: therefore (I. 8.) the angle DEF is equal to GEF, DFE to GFE, and EDF to ÉGF. Then, because the angle DEF is equal to GEF, and (const.) GEF to ABC; therefore the angle ABC is equal to DEF. For the same reason, ACB is equal to DFE, and A to D. Therefore the triangles ABC, DEF are equiangular : wherefore, if the sides, &c.

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PROP. VI. THEOR.

If two triangles have one angle of the one equal to one angle of the other, and the sides about the equal angles proportionals; the remaining angles are equal, each to each, viz., those which are opposite to the homologous sides.

Let the triangles ABC, DEF have the angles BAC, EDF equal, and the sides about those angles proportionals ; that is, BA: AC :: ED: DF; the angle ABC is equal to DEF, and ACB to DFE.

Make (1. 23.) the angle FDG equal to either of the angles BAC, EDF; and the angle DFG equal to ACB. Then (VI. 4. cor.)

BA : AC :: GD: DF. But (hyp.)

BA : AC::ED: DF; and therefore (V. 11.) ED: DF::GD: DF; wherefore ED is equal (V. 9.) to DG. Now DF is common to the two triangles EDF, GDF; and the angle EDF is equal (const.) to GDF; therefore the angle DFG is equal (I. 4. part 3.) to DFE, and G to E. But (const.) the angle DFG is equal to ACB: therefore ACB is equal to DFE; and (hyp.) the angle BAC is equal to EDF; wherefore also (I. 32. cor. 5.) the remaining angles B and E are equal. Therefore, if two triangles, &c.

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OTHERWISE. Let ABC, ADE be the triangles, having the angles at A equal, and BA: AC::DA: AE. Then the homologous sides BA, AD being placed in the same straight line, and contiguous, and the triangles being placed on opposite sides of BD, the sides AC, AE are (I. 14. cor.) in the same straight line. Also (VI. 2. cor. part. 2.) BC and DE are parallel; and therefore (I. 29.) the angles B and D are equal, and likewise C and E. .

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PROP. VII. THEOR. If two triangles have two sides of the one proportional to two sides of the other; and if the angles opposite to one pair of the homologous sides be equal, and those opposite to the other pair be either both acute, or not acute; the angles contained by the proportional sides are equal.

In the triangles ABC, DEF, let AB : BC :: DE: EF, and let the angles A and D be equal : then if C and F be either both acute, or not acute, the angles ABC, DEF are equal.

For, if they be not equal, let ABC be the greater, and (I. 23.) make the angle ABG equal to DEF: then, since (hyp.) A and D are equal, AGB is equal (1. 32. cor. 5.) to DFE; and (VI. 4.)

AB : BG :: DE: EF; but (hyp.)

AB : BC::DE: EF: therefore (V. 11.) AB : BG :: AB : BC. Hence (V. 9.) BG and BC are equal, because AB has the same ratio to each of them; and therefore (I. 5.) the angle BGC is equal to C. The angles at G therefore are both acute, or not acute, according as C and F are acute, or not acute.

In the first case, then, in which C and F are both acute, the angles BGA, BGC are together less than two right angles, which (I. 13.) is impossible ; and in the second case, in which C and F are not acute, the angles BGC, BCG of the triangle CBG are not together less than two right angles, which (I. 17.) is impossible. The angle ABC, therefore, is not unequal to DÉF; that is, it is equal to it: wherefore, if two triangles, &c. +

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* This proposition admits of a very easy indirect proof.

+ The proof of this proposition which is given above, is much shorter than that of Euclid. The expression of the enunciation is also altered, with a view to render it more simple and plain,

PROP. VIII. THEOR.

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In a right-angled triangle, if a perpendicular be drawn from the right angle to the hypotenuse, 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 hypotenuse BC: the triangles A DB, ADC are similar to the whole triangle ABC, and to one another.

Because the angle BAC is equal (I. ax. 11.) to ADB, each of them being a right angle, and that the angle B is common to the two triangles ABC, ABD; the remaining angle C is equal (I. 32. cor. 5.) to the remaining angle BAD. Therefore the triangles ABC, ABD are equiangular, and (VI. 4.) the sides about their equal angles are proportionals ; wherefore (VI. def. 1.) the triangles are similar. In the same manner it might be demonstrated, that the triangle ADC is equiangular and similar to ABC: and the triangles ADB, ADC, being each equiangular to ABC, are equiangular, and therefore (VI. 4. and def. 1.) similar, to each other. Therefore, in a right-angled triangle, &c.

Cor. From this it is manifest, that the perpendicular drawn from the right angle of a right-angled triangle to the hypotenuse, is a mean proportional (V. def. 9.) between the segments of the hypotenuse : and also that each of the sides is a mean proportional between the hypotenuse and its segments adjacent to that side. For, (VI. 4.) in the triangles BDA, ADC,

as BD : DA :: DA: DC; in the triangles ABC, DBA, as BC: BA :: BA : BD ; and, in the triangles ABC, ÁCD, as BC : CA::CA: CD.

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From a given straight line to cut off any assigned part, or submultiple.

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

* This problem is the same in substance, as to divide a given straight line into any proposed number of equal parts ; a method of doing which has already been given in the second corollary to the 34th proposition of the first book.

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