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PROPOSITION 36. THEOREM. Parallelograms on equal bases, and between the same parallels, are equal to one another.

Let ABCD, EFGH be parallelograms on equal bases BC, FG, and between the same parallels AH, BG: the parallelogram ABCD shall be equal to the parallelogram EFGH.

Join BE, CH.

Then, because BC is equal to FG,[Hyp. and FG to EH,[I. 34. BC is equal to

[Axiom 1. and they are parallels,

[Hypothesis. and joined towards the same parts by the straight lines BE, CH. But straight lines which join the extremities of equal and parallel straight lines towards the same parts are themselves equal and parallel.

[I. 33. Therefore BE, CH are both equal and parallel. Therefore EBCH is a parallelogram.

[Definition. And it is equal to ABCD, because they are on the same base BC, and between the same parallels BC,AH. [I. 35.

For the same reason the parallelogram EFGH is equal to the same EBCH. Therefore the parallelogram ABCD is equal to the parallelogram EFGH.

[Axiom 1. Wherefore, parallelograms &c. Q.E.D.

PROPOSITION 37. THEOREM. Triangles on the same base, and between the same par. allels, are equal.

Let the triangles ABC, DBC be on the same base BC, and between the same parallels AD, BC: the triangle ABC shall be equal to the triangle DBC.

Produce AD both ways to the points E, F; [Post. 2.

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through B draw BE parallel to CA, and through C draw CF parallel to BD.

[1. 31. Then each of the figures EBCA, DBCF is a parallelogram;

[Definition and EBCA is equal to DBCF, because they are on the same base BC, and between the same parallels BC, EF. [I. 35. And the triangle ABC is half of the parallelogram EBCA, because the diameter AB bisects the parallelogram; [I. 34. and the triangle DBC is half of the parallelogram DBCF, because the diameter DC bisects the parallelogram. [I. 34. But the halves of equal things are equal. [Axiom 7. Therefore the triangle ABC is equal to the triangle DBC. Wherefore, triangles &c. Q.E.D.

PROPOSITION 38. THEOREM. Triangles on equal bases, and between the same parallels, are equal to one another.

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

Produce ADboth ways to the points G, H; through B draw BG parallel to CA, and through F draw FH parallel to ED. [I. 31.

Then each of the figures GBCA, DEFH is a parallelogram. [Definition. And they are equal to one another because they are on equal bases BC, EF, and between the same parallels

[I. 36. And the triangle ABC is half of the parallelogram GBCA, because the diameter AB bisects the parallelogram; [I. 34. and the triangle DEF is half of the parallelogram DEFH, because the diameter DF bisects the parallelogram. But the halves of equal things are equal. [Axiom 7. Therefore the triangle ABC is equal to the triangle DEF.

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

B

BF, GH.

B

PROPOSITION 39. THEOREM. Equal triangles on the same base, and on the same side of it, are between the same parallels.

Let the equal triangles ABC, DBC be on the same base BC, and on the same side of it: they shall be between the same parallels.

Join AD.
AD shall be parallel to BC.

E
For if it is not, through A draw
AE parallel to BC, meeting BD
at E.

[I. 31. and join EC. Then the triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE.

[I. 37. But the triangle ABC is equal to the triangle DBC. (Hyp. Therefore also the triangle DBC is equal to the triangle EBC,

[Axiom 1. the greater to the less; which is impossible. Therefore AE is not parallel to BC.

In the same manner it can be shewn, that no other straight line through A but AD is parallel to BC; therefore AD is parallel to BC.

Wherefore, equal triangles &c. Q.E.D.

PROPOSITION 40. THEOREM, Equal triangles, on equal bases, in the same straight line, and on the same side of it, are between the same parallels.

Let the equal triangles ABC, DEF be on equal bases BC, EF, in the same straight line BF, and on the same side of it: they shall be between the same parallels.

Join AD.
AD shall be parallel to BF.

For if it is not, through A draw AG parallel to BF, meeting ED at G [I. 31. and join GF

B

Then the triangle ABC is equal to the triangle GEF, because they are on equal bases BC, EF, and between the same parallels.

[I. 38. But the triangle ABC is equal to the triangle DEF. (Hyp. Therefore also the triangle DEF is equal to the triangle GEF,

[Axiom 1 the greater to the less; which is impossible. Therefore AG is not parallel to BF.

In the same manner it can be shown that no other straight line through A but AD is parallel to BF; therefore AD is parallel to BF.

Wherefore, equal triangles &c. Q.E.D.

PROPOSITION 41. THEOREM.

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If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle.

Let the parallelogram ABCD and the triangle EBC be on the same base BC, and between the same parallels BC, AE: the parallelogram ABCD shall be double of the triangle EBC.

Join AC.

Then the triangle ABC is equal to the triangle EBC, because they are on the same base BC, and between the same parallels BC, AE. [I. 37. But the parallelogram ABCD is double of the triangle ABC, because the diameter AC bisects the parallelogram. [I. 34.

Therefore the parallelogram ABCD is also double of the triangle EBC.

Wherefore, if a parallelogram &c. Q.E.D.

B

PROPOSITION 42. PROBLEM.

G

to D;

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To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

Let ABC be the given triangle, and D the given rectilineal angle: it is required to describe a parallelogram that shall be equal to the given triangle ABC, and have one of its angles equal to D.

Bisect BC at E:[1. 10. join AE, and at the point E, in the straight line EC, make the angle CEF equal

[I. 23. through A draw AFG parallel to EC and through Ĉ draw CG parallel to EF.

[I. 31. Therefore FECG is a parallelogram.

[Definition. And, because BE is equal to EC, [Construction. the triangle ABE is equal to the triangle AEC, because they are on equal bases BE, EC, and between the same parallels BC, AG.

[I. 38. Therefore the triangle ABC is double of the triangle A EC. But the parallelogram FECG is also double of the triangle A EC, because they are on the same base EC, and between the same parallels EC, AG.

[1. 41. Therefore the parallelogram FECG is equal to the triangle ABC;

[Axiom 6. and it has one of its angles CEF equal to the given angle

[Construction. Wherefore a parallelogram FECG has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle D. Q.E.F.

D.

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