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to each of its angles. And, by the preceeding proposition, all

the angles of these triangles are equal to twice as many right.

angles as there are triangles, that is, as there are fides of the figure; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles; that is", together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal to twice as many right angles as the figure has fides.

CoR. 2. All the exterior angles of any reëtilineal figure, are together equal to four right angles.

Because every interior angle ABC, with its adjacent exterior ABD, is equal b to two right angles; therefore all the interi- A. or, together with all the exterior angles of the figure, are equal to twice as many right angles as there are fides of the figure; that is, by the foregoing corollary, they are equal to all the inte- D B rior angles of the figure, toge- ther with four right angles; therefore all the exterior angles are equal to four right angles.

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THE opposite sides and angles of parallelograms are equal to one another, and the diameter bisects them,

that is, divides them in two equal parts.

N. B. A parallelogram is a four sided figure, of which the opposite sides are parallel; and the diameter is the

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Let ACDB be a parallelogram, of which BC is a diameter; the opposite sides and angles of the figure are equal to one another; and the diameter BC bisects it.

Because AB is parallel to CD, A B and BC meets them, the alternate angles ABC, BCD are equal * to one another ; and be- a 29. I, cause AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD, are equal * to one another; wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other, each to each, and one side BC common to the two triangles, which is adjacent to their equal angles ; therefore their other fides shall be equal, each to each, and the third angle of the one to the third angle of the other", viz. b 26, 1, the fide AB to the fide CD, and AC to BD, and the angle BAC equal to the angle BDC : And because the angle ABC is equal to the angle BCD, and the angle CBD to the angle ACB; the whole angle ABD is equal to the whole angle ACD : And the angle BAC has been shown to be equal to the angle BDC; therefore the opposite sides and angles of parallelograms are equal to one another ; also, their diameter bisects them; for AB being equal to CD, and BC common, the two AB, BC are equal to the two DC, co to each; and the angle ABC

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is equal to the angle BCD ; therefore the triangle ABC is equal" to the triangle BCD, and the diameter BC divides the parallelogram ACDB into two equal parts. Q. E. D.

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ARALLELog RAMs upon the same base and between the same parallels, are equal to one another.

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the two FID, DC, each to each; and the exterior angle FDC is equal" to the interior EAB; therefore the base EB is equal to the base FC, and the triangle EAB equal “to the triangle FDC; take the triangle FDC from the trapezium ABCF, and

from the same trapezium take the triangle EAB; the remain-,

ders therefore are equal f, that is, the parallelogram ABCD is equal to the parallelogram EBCF. Therefore parallelograms upon the same base, &c. Q. E. D.

f 3. Ax.

P R O P.

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ARALLELoGRAMs upon equal bases and between the same parallels, are equal to one another.

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a 31. Ia

the half of the parallelogram EBCA, because the diameter AB bise&ts it; and the triangle DBC is the half of the parallelogram DBCF, because the diameter DC bise&s it: But the halves of equal things are equal"; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c. Q. E. D.

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RIANGLEs upon equal bases, and between the same parallels, are equal to one another.

Let the triangles ABC, DEF be upon equal bases BC, EF, and between the same parallels BF, AD: The triangle ABC is equal to the triangle DEF.

Produce AD both ways to the points G, H, and through B draw BG o CA, and through F draw FH parallel to ED : Then each o the figures GBCA, G. A. D H DEFH is a parallelogram; and they are equal"to one another, because they are upon equal bases BC, EF and between the same parallels B C E F BF, GH; and the triangle ABC is the half of the parallelogram GBCA, because the diameter AB bise&ts it; and the triangle DEF is the half “ of the parallelogram DEFH, because the diameter DFbise&ts it: But the halves of equal things are equal"; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c. Q. E. D.

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Let the equal triangles ABC, DBC be upon the same base
BC, and upon the same fide of it; they are between the same

parallels.
Join AD; AD is parallel to BC; for, if it is not, through
the point A draw "AE parallel to BC, and join EC: The tri-
angle

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