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PROPOSITION XXX.

THEOREM.
Right Lines parallel to one and the fame Right

Line, are also parallel between themselves.
LET A B and CD be Right Lines, each of which

is parallel to the Right Line E F. I say, AB is also parallel to CD. For let the Right Line K fail

upon them.

Then because the Right Line GK falls upon the * 29 of 1 bis. parallel Right Lines AB, EF, the Angle AG H is *

equal to the Angle GHF; and because the Right

Line GK falls upon the parallel Right Lines EF, + 29 of tbis. C D, the Angle G H F is equal to the Angle GKD+

But it has been proved that the Angle A GK is also equal to the Angle G HF. Therefore 'A G K is

equal GKD, and they are alternate Angles; whence 1 27 of ibis. A B is parallel to CD I. And so, Right Lines, pa

rallel to one and the same Right Lines, are parallel between themselves; which was to be demonstrated.

PROPOSITION XXXI.

PROBLEM.
To draw a Right Line through a given Point

parallel to a given Right Line.
L ET A be a Point given, and B C a Right Line

given. It is required to draw a Right Line thro the Point A, parallel to the Right Line B C.

Assume any Poine D in B C, and join AD; then • 23 of tbis. make * an Angle D A E, at the Point A, with the

Li DA, equal to the Angle ADC, and produce
E A strait forwards to F.

Then, because the Right Line A D, falling on two
Right Lines B C, EF, makes the alternate Angles

EAD, AD C, equal between themselves, E F fball +27 of «bis, be + parallel to BC. Therefore, the Right Line

E AF, is drawn from the given Point A, parallel to the given Right Line BC; which was to be done,

PRO

PROPOSITION XXXII.

THEOREM.
If one side of any Triangle be produced, the out-

ward Angle is equal to both ibe inward and op-
posite Angles: and the three inward Angles of

a Triangle are equal to two Right Angles. L ET A B C be a Triangle, one of whose Sides

B C is produced to D. I say the outward Angle ACD is equal to the two inward and opposite Angle C A B, A B C; and the three inward Angles of the Triangle, viz. A B C, BCA, C AB, are equal to two Right Angles. For let CE be drawn * thro' the Point C, parallel.

31 of this to the Right Line A B. Then, because A B is parallel to CE, and A C falls upon them, the alternate Angles BAC, ACE, are + equal between themselves. † 29 of sbis. Again, because A B is parallel to CE, and the Right Line B D falls upon them, the outward Angle ECD is || equal to the inward and opposite one A B C; but || 29 of thisu it has been proved that the Angle ACE is equal to the Angle B AC. Wherefore the whole outward Angle A CD is equal to both the inward and opposite Angles B AC, A BC. And if the Angle A CB, which is common, be added, the two Angles A CD, ACB, are equal to the three Angles A B C, B AC, ACB; but the Angles A CD, AC B, are I equal to 1 13 of ibido two Right Angles. Therefore also shall the Angles ACB, CBA, CAB, be equal to two Right Angles. Wherefore, If one side of any Triangle be produced, the outward Angle is equal to both the inward and opposite Angles, and the three inward Angles of a Triangle are equal 10 two Right Angles ; which was to be demonstrated. Coroll. 1. All the three Angles of any one Triangle

taken together, are equal to all the three Angles of any other Triangle, taken together. Caroll. 2. If two Angles of any one Triangle, either

separately, or taken together, be equal to two Angles of any other Triangle; then the remaining Angle

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of the one Triangle will be equal to the remaining 'Angle of the other. Coroll. 3. If one Angle of a Triangle be a Right An

gle, the other two Angles together make one

Right Angle. Coroll. 4. If the Angle included between the equal

Legs of an Isoceles Triangle be a Right one, each of the other Angles at the Base will be half a Right

Angle. Coroll. 5. Any Angle in an Equilateral Triangle is

equal to one Third of two Right Angles, or two

Thirds of one Right Angle. Coroll. 6. Hence it appears, that if one Angle of any

Triangle be equal to the other two, that is a Right one ; because that the Angle adjacent to this Right one, is equal to the other two. But when adjacent Angles are equal, they are necessarily Right ones.

THEOREM I.

All the inward Angles of any Right-lind Fi

gure whatsoever, make twice as many Right Angles, abating four, as the Figure has Sides.

FOR

any Right-lin'd Figure may be resolved into as many Triangles, abating two, as it hath Sides. For Example, if a Figure has four Sides, it may be refolved into two Triangles : If a Figure hath five Sides, it may be resolved into three Triangles; if fix, into four ; and so on. Wherefore (by Prop. XXXII.) the Angles of all these Triangles are equal to twice as many Right Angles, as there are Triangles : But the Angles of all the Triangles are equal to the inward Angles of the Figure. Therefore all the inward Angles of the Figure are equal to twice as many Right Angles as there are Triangles, that is, twice as many Right Angles, taking away four, as the Figure hath Sides. WWD.

THE

THEOREM II.
All the outward Angles of any Right-lin'd Fi.

gure, together make four Right Angles. FOR

OR the outward Angles, together with the inward

ones, make twice as many Right Angles as the Figure has Sides; but from the last Theorem, all the inward Angles, together, make twice as many Right Angles, abating four, as the Figure hath Sides. Wherefore the outward Angles are, all together, equal to four Right Angles. W. W.D.

PROPOSITION XXXIII.

THEOREM.
Two Right Lines which join two equal and pa-

rallel Rigbt Lines towards the same Parts, are
also equal and parallel.

E T the parallel and equal Right Lines A B, CD,

be joined, towards the fame Parts, by the Right Lines AC, BD. I say, A C, B D, are equal and parallel.

For draw BC.

Then, because A B is parallel to CD, and B C falls upon them, the alternate Angles ABC, BCD, are * equal. Again, because A B is equal to Ç D, and * 29 of tbis. BC is common; the two Sides A B, BC, are equal to the two Sides, BC, CD; but the Angle A B C is also equal to the Angle BCD; therefore ihe Base AC is + equal to the Bale BD: And the Triangle ABC + 4 of tbisa equal to the Triangle BCD; and the remaining Angles equal to the remaining Angles, each to each, which subtend the equal Sides. Wherefore the Angle ACB is equal to the Angle CBD. And becaule the Right Line BC, falling upon two Right Lines AC, BD, makes the alternate Angles ACB, CBD, equal between themselves ; A C is 1 parallel to B D. But it t 29 of 1b:s. has been proved also to be equal to it. Therefore, two Right Lines, which join two equal and parallel Right Lines, towards the fame Parts, are also equal and parallel ; which was to be demonstrated. D

Defin.

Defin. A Parallelogram is a Quadrilateral Figure, each

of whose opposite Sides are parallel.

PROPOSITION XXXIV.

THEOREM.
The opposite Sides and opposite Angles of any Pa.

rallelogram are equal ; and the Diameter di

vides tbe fame into two equal Parts. LET ABCD be a Parallelogram, whose Diame

ter is BC. I say, the opposite sides and opposite Angles are equal between themselves, and the Diameter B C bisects the Parallelogram.

For, because A B is parallel to CD, and the Right

Line B C falls on them, the alternate Angles A B C, • 29 of ibis. BCD, are * equal between themselves ; again, be

cause AC is parallel to B D, and B C falls upon them, the alternate Angles ACB, and CBD, are equal to one another. Wherefore ABC, CBD, are two Triangles, having two Angles ABC, BCÁ, of the one, equal to two Angles B C D, CBD, of the other, each to each ; and likewise one side of the one equal to one side of the other, viz. the Side B C between the equal Angles, which is common.

There+ 26 of ikis, fore the remaining Sides shall be + equal to the remain

ing Sides, each to each, and the remaining Angle to the remaining Angle. And so the Side A B is equal to the Side CD, the Side A C to BD, and the Angle BAC to the Angle B D C. And because the Angle ABC is equal to the Angle B CD, and the Angle CBD to the Angle ACB; therefore the whole Angle ABD is equal to the whole Angle A CD: But it has been proved that the Angle BAC is also equal to the Angle Ᏼ Ꭰ C.

Wherefore, the opposite Sides and Angles of any Parallelogram are equal between themselves.

I fay, moreover, that the Diameter bisects it. For because A B. is equal to CD, and BC is common, the two Sides A B, BC, are each equal to the two

Sides DC, CB; and the Angle A B C is also equal * 4 of rbise to the Angle DCB. Therefore the Base A C is I

equal

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