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 An-
gles 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 Ifofceles Triangle be a Right one, each
of the other Angles at the Bafe 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.

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
OR any Right-lined Figure may be refolved into
as many Triangles, abating two, as it hath Sides.
For Example, if a Figure has four Sides, it may be re-
folved into two Triangles: If a Figure has five Sides,
it may be refolved into three Triangles; if fix, into
four; and fo on. Wherefore (by Prop. XXXII.) the
Angles of all thefe 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 has 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.

THE ORE M.

Two Right Lines, which join two equal and parallel Right Lines, towards the fame Parts, are alfo equal and parallel.

ET the parallel and equal Right Lines AB, CD, be joined towards the fame Parts, by the Right Lines AC, BD. I fay 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 alfo equal to the Angle BCD; therefore the + 4 of this. Bafe AC is † equal to the Bafe 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 because the Right Line BC, falling upon two Right, 127 of this. Lines AC, BD, makes the alternate Angles ACB, CBD, equal between themselves; AC is ‡ parallel to BD. But it has been proved alfo 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.

Defin. A Parallelogram is a Quadrilateral Figure, each of whofe oppofite Sides are parallel.

PROPOSITION XXXIV.

THEOREM.

The oppofite Sides and oppofite Angles of any Parallelogram are equal; and the Diameter divides the fame into two equal Parts.

LET ABC. Lay,

ET ABDC be a Parallelogram, whofe Diameter is BC. I fay, the oppofite Sides and opposite Angles are equal between themfelves, and the Diameter BC bifects the Parallelogram.

*

For because AB is parallel to CD, and the Right Line BC falls on them, the alternate Angles ABC, BCD, are equal between themselves; again, be- * 29 of this caufe 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 likewife one Side of the one equal to one Side of the other, viz. the Side BC between the equal Angles, which is common. Therefore the remaining Sides fhall be † equal to the remaining Sides, † 26 of this each to each, and the remaining Angle to the remaining Angle. And fo 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 oppofite Sides and Angles of any Parallelogram are equal between themselves.

I fay, moreover, that the Diameter bifects 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 alfo equal to the Angle BCD. Therefore the Bafe AC is equal to the Bafe DB; and the Triangle ABC is 14 of this equal

Ax. 2.

equal to the Triangle BCD. Wherefore the Diameter BC bifects the Parallelogram ACDB; which was to be demonftrated.

PROPOSITION XXXV.

THEOREM.

Parallelograms conftituted upon the fame Bafe, and between the fame Parallels, are equal between themselves.

LE

ET ABCD, EBCF, be Parallelograms conftituted upon the fame Base BC, and between the fame Parallels AF and BC. I fay, the Parallelogram ABCD, is equal to the Parallelogram EBCF.

For because ABCD is a Parallelogram, AD is 34 of this equal to BC; and for the fame Reason EF is equal Axiom 1 to BC; wherefore AD fhall be equal to EF; but DE is common. Therefore the whole A E is 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 this. and the Angle FDC* equal to the Angle EA B, the outward one to the inward one. Therefore the Base + 4 of this. EB is equal to the Bafe CF, and the Triangle EAB to the Triangle FDC. If the common Triangle DGE be taken from both, there will remain 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 conftituted upon the fame Bafe, and between the fame Parallels, are equal between themselves; which was to be demonftrated.

4x. 3.

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