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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. Coroll

. 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 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 Angle, the other two Angles together make one

Right Angle. Coroll

. 4. If the Angle included between the equal Legs of an Isosceles Triangle be a Right one, each of the other Angles at the Base will be half Right

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

equal to one Third of two Right Angles, or two
Thirds of one Right Angle.

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THEOREM I.

All the inward Angles of any Right-lined Fi

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

Sides.
FOR any Right-lined 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 has 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 has Sides. W. W.D.

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THEOREM II. All the outward Angles of any Right-lined Figure together, make four Right

Angles. FOR the outward Angles, together with the inward

ones, make twice as many Right Angles as the Figure bas Sides; but from the last Theorem, all the inward Angles together make twice as many Right Angles, abating four, as the Figure has Sides. Wherefore the outward

Angles are all together equal to four Right Angles. W.W.D.

PROPOSITION XXXII.

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

rallel Right Lines, towards the same Parts,
are also equal and parallel.
ET the parallel and equal Right Lines A B, CD,

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

For draw BC.

Then because AB is parallel to CD, and BC falls

upon them, the alternate Angles ABC, BCD, are 29 of this. * equal. Again, because AB is equal to CD, and

BC is common; the two Sides AB, BC, are each equal to the two Sides BC, CD; but the Angle

ABC is also equal to the Angle BCD; therefore the + 4 of this. Base AC is equal to the Base BD: And the Triangle

ABC, equal to the Triangle BCD; and the remaining Angles, equal to the remaining Angles, each to each, which fubtend the equal Sides. Wherefore the Angle ACB is equal to the Angle CBD. And be

cause the Right Line BC, falling upon two Right; 127 of tbis. Lines AC, BD, makes I the alternate Angles ACB,

CBD, equal between themselves ; AC is 1 párallel to BÓ. But it has been proved also to be equal to it. Therefore two Right Lines, which join two equal and parallel Right Lines, towards the same Parts, are also equal and parallel ; which was to be demonstrated.

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 ibe same into two equal Parts.

ET ABDC 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 AB is parallel to CD, and the Right Line B C falls on them, the alternate Angles ABC, BCD, are * equal between themselves; again, be- * 29 of this cause AC is parallel to BD, and BC falls upon them, the alternate Angles ACB and CBD are equal to one another, Wherefore ABC, CBD, are two Triangles, having two Angles ABC, BCA, of the one, equal to two Angles BCD, 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. Therefore the remaining Sides shall be † equal to the remaining Sides, † 26 of this each to each, and the remaining Angle to the remaining Angle. And so the Side AB is equal to the Side CD), the Side AC to BD, and the Angle BAC to the Angle BDC. And because the Angle ABC is .equal to the Angle BCD, and the Angle CBD to the Angle ACB; therefore the whole Angle ABD is equal to the whole Angle ACD: But it has been proved, that the Angle BAC is also equal to the Angle BDC.

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

I say, moreover, that the Diameter bisects it. For because AB is equal to CD, and BC is common, the two Sides AB, BC, are each equal to the two Sides DC, CB; and the Angle ABC is also equal to the Angle BCD. Therefore the Base AC is I equal to the Bale DB; and the Triangle ABC is I 4 of this

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equal to the Triangle BCD. Wherefore the Diameter BC bisects the Parallelogram ACDB; which was to be demonstrated.

PROPOSITION XXXV.

THEOREM.
Parallelograms conftituted upon the same Base,

and between the same Parallels, are equal be

tween themselves. Le

ET ABCD, EBCF, be Parallelograms conthe fame Parallels AF and BC. I say, the Parallelogram ABCD, is equal to the Parallelogram EBCF.

For because ABCD is a Parallelogram, AD is * 34 of tbis. * equal to BC; and for the same Reason EF is equal 4 Axiom 1. to BC; wherefore AD shall be equal to EF; but | Ax. 2. DE is common. Therefore the whole A E is I equal

to the whole DF. But AB is equal to DC; wherefore EA, AB, the two sides of the Triangle ABE,

are equal to the two Sides FD, DC, each to each ; * 29 of tbis, and the Angle FDC* equal to the Angle EAB, the

outward one to the inward one. Therefore the Base t 4 of tbi:. E B is t equal to the Base CF, and the Triangle EAB

to the Triangle FDC. If the common Triangle Ax. 3. DGE be taken from both, there will remain I the

Trapezium ABGD, equal to the Trapezium FCGE; and if the Triangle GBC, which is common, be added, the Parallelogram ABCD will be equal to the Parallelogram EBCF. Therefore, Parallelograms constituted upon the fame Bafe, and between the same Parallels, are equal between themselves ; which was to be demonstrated.

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Prop. 33

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