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as the base BC is to the base CD; therefore (5. 11), as the base BC is to the base CD, so is the parallelogram EC to the parallelogram CF.

Wherefore, Triangles &c. Q. E.D.

COR. Triangles and parallelograms of equal altitudes are to one another as their bases.

PROP. II. THEOR.

If a straight line be parallel to the base of a triangle, it shall cut the sides, or the sides produced, proportionally: and if the sides, or the sides produced, be cut proportionally, the straight line which joins the points of section shall be parallel to the base.

Let DE be parallel to BC, the base of the triangle ABC: BD shall be to DA as CE to EA.

And

A

FB

B D

A

Join BE, CD: Then the triangle BDE is equal to the triangle CDE, because they are on the same base DE, and between the same parallels DE, BC: ADE is another triangle; therefore (5. 7) the triangle D BDE is to the triangle ADE as the triangle CDE is to the B triangle ADE: But the triangle BDE is to the triangle ADE as BD is to DA (6. 1), because they have the same altitude, viz. the perpendicular from E on AB; and so also the triangle CDE is to the triangle ADE as CE to EA: Therefore BD is to DA as CE to EA.

C D

E B

Next, let BD be to DA as CE to EA, and join DE: then DE shall be parallel to BC.

For, the same construction being made, because BD is to DA as CE to EA, and as BD to DA so is the triangle BDE to the triangle ADE, and as CE to EA so is

the triangle CDE to the triangle ADE--therefore the triangle BDE is to the triangle ADE as the triangle CDE to the triangle ADE, and therefore (5.9) the triangle BDE is equal to the triangle CDE: But they are on the same base DE; therefore DE is parallel to BC.

Wherefore, If a straight line &c. Q. E. D.

PROP. III. THEOR.

If the vertical angle of a triangle be bisected by a straight line which also cuts the base, the segments of the base shall have the same ratio which the other sides of the triangle have to one another: and if the segments of the base have the same ratio which the other sides of the triangle have to one another, the straight line, drawn from the vertex to the point of section, bisects the vertical angle.

Let the vertical angle BAC of any triangle ABC be bisected by the straight line AD, which cuts the base BC in D: BD shall be to DC as BA to AC.

Through C draw CE parallel to AD, meeting BA produced in E: Then, because AC meets the two parallels AD, EC, the angle DAC is equal to the alternate angle ACE: But the angle DAC is equal to the angle BAD; therefore also the angle

BAD is equal to the angle ACE:
Again, because BAE meets the two
parallels AD, EC, the exterior angle
BAD is equal to the interior and op-

E

posite angle AEC: But the angle BAD has been proved equal to the angle ACE; therefore also the angle ACE is equal to the angle AEC, and the side AC to the side AE: Now because AD is parallel to CE,

a side of the triangle BCE, therefore (6. 2) BD is to DC as BA to AE: But AE is equal to AC; therefore BD is to DC as BA to AC.

Next, let BD be to DC as BA to AC, and join AD: then the angle BAC shall be bisected by AD.

For, the same construction being made, because BD is to DC as BA to AC, and also BD is to DC as BA to AE (since AD is parallel to EC), therefore BA is to AC as BA to AE: Therefore (5. 9) AC is equal to AE and the angle ACE to the angle AEC: But the angle ACE is equal to the alternate angle DAC, and the angle AEC is equal to the exterior and opposite angle BAD; therefore the angle BAD is equal to the angle DAC, that is, the angle BAC is bisected by the straight line AD.

Wherefore, If the vertical angle &c. Q.E.D.

PROP. A. THEOR.

If the exterior angle of a triangle, made by producing any one of its sides, be bisected by a straight line which also cuts the base produced, the segments between the dividing line and the extremities of the base shall have the same ratio which the other sides of the triangle have to one another: and if the segments of the base produced have the same ratio which the other sides of the triangle have to one another, the straight line drawn from the vertex to the point of section shall bisect the exterior angle of the triangle.

Let the exterior angle CAE of any triangle ABC be bisected by AD, which meets the base produced in D: then BD shall be to DC as BA to AC. Through C draw CF parallel to AD: AC meets the two parallels AD, CF, is equal to the alternate angle ACF:

Then, because the angle DAC But the angle

E

DAC is equal to the angle DAE; therefore also the angle DAE is equal to the angle ACF: Again, because FAE meets the two parallels AD, FC, the exterior angle DAE is equal to the interior and opposite angle AFC: But the angle

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C

DAE was proved equal to the angle ACF; therefore also the angle ACF is equal to the angle AFC, and the side AC to the side AF: Now because AD is parallel to CF, a side of the triangle BCF, therefore BD is to DC as BA to AF: But AF is equal to AC; therefore BD is to DC as BA to AC.

Next, let BD be to DC as BA to AC, and join AD : then the angle CAE shall be bisected by AD.

For, the same construction being made, because BD is to DC as BA to AC, and also BD is to DC as BA to AF, therefore BA is to AC as BA to AF, and therefore AC is equal to AF, and the angle ACF to the angle AFC: But the angle ACF is equal to the alternate angle CAD, and the angle AFC to the exterior angle DAE; therefore also the angle CAD is equal to the angle DAE, that is, the angle CAE is bisected by AD. Wherefore, If the exterior angle &c.

PROP. IV. THEOR.

Q. E.D.

The sides about the equal angles of equiangular triangles are proportionals, and those 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 the angle DCE, and the angle ACB to the angle DEC, and consequently the angle BAC equal to the angle CDE: then the sides about the equal

angles of the triangles, ABC, DCE, shall be proportionals, those being the homologous sides F which are opposite to the equal angles.

A

B

D

Let the triangle DCE be so placed that its side CE may be contiguous to BC, and in the same straight line with it Then, because the angles ABC, ACB, are together less than two right angles, and that the angle ACB is equal to the angle DEC, therefore also the angles ABC, DEC are less than two right angles, and therefore BA, ED, if produced, shall meet; let them be produced and meet in F: Then, because the angle ABC is equal to the angle DCE, therefore BF is parallel to CD; and, again, because the angle ACB is equal to the angle DEC, therefore AC is parallel to FE: Therefore FACD is a parallelogram, and AF is equal to CD, and AC to FD: Now, because AC is parallel to FE, one of the sides of the triangle FBE, therefore BA is to AF as BC to CE: but AF is equal to CD; therefore BA is to CD as BC to CE, and, alternately (5. 16), AB is to BC as DC to CE: Again, because CD is parallel to BF, therefore BC is to CE as FD to DE: but FD is equal to AC; therefore BC is to CE as AC to DE, and, alternately, BC is to CA as CE to ED: Lastly, because it has been proved that AB is to BC as DC to CE, and BC to CA as CE to ED, therefore, ex æquali (5. 22), BA is to AC as CD to DE.

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If the sides of two triangles, about each of their angles, be proportionals, the triangles shall be equiangular, and have those angles equal which are opposite to homologous sides.

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