PROP. XXXVI. THEOREM.

Parallelograms on equal bases and between the same
parallels are equal to one another.

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

Then it is to be proved that

The parallelogram ABCD the parallelogram EFGH.

CONSTRUCTION.-Join BE and CH.

PROOF.-Because_BC= FG (hyp.), and because FG = EH (I. 34), therefore BC = EH (ax. 1); also, because BC and EH are parallels and joined towards the same parts by BE and CH (hyp.), therefore BE and CH are both equal and parallel (I. 33); and therefore the figure EBCH is a parallelogram (def. 34).

Next, because the parallelograms ABCD and EBCH are upon the same base BC and between the same parallels AH and BC, therefore the parallelogram ABCD = the parallelogram EBCH (I. 35).

Similarly, the parallelogram EFGH = the parallelogram EBCH, being upon the same base EH, and between the same parallels EH and BG.

Therefore, it is proved, as required, that (ax. 1)

The parallelogram ABCD=the parallelogram EFGH. Wherefore,

Parallelograms on equal bases, &c.

Q. E. D.

PROP. XXXVII. THEOREM.

Triangles on the same base and between the same
parallels are equal to one another.

Let ABC and DBC be triangles on the same base BC, and between the same parallels AD and BC.

CONSTRUCTION.-1. Produce AD both ways to points

E and F.

2. Through B draw BE parallel to CA, and through C draw CF parallel to BD (I. 31).

PROOF.-Because each of the figures EBCA and DBCF is a parallelogram (def. 34), and because they are on the same base BC, and between the same parallels BC and EF, therefore the parallelogram EBCA = the parallelogram DBCF (I. 35).

Next, because the triangle ABC is half the parallelogram EBCA, and the triangle DBC is half the parallelogram DBCF (I. 34),

Therefore, it is proved, as required, that

The triangle ABC the triangle DBC (ax. 7),

Wherefore,

Triangles on the same base, &c.

Q. E. D.

PROP. XXXVIII. THEOREM.

Triangles on equal bases and between the same parallels are equal to one another.

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

Then it is to be proved that

The triangle ABC the triangle DEF.

CONSTRUCTION.-1. Produce AD both ways to G and H. 2. Through B draw BG parallel to CA, and through F draw FH parallel to ED (I. 31).

PROOF.-Because each of the figures GBCA and DEFH is a parallelogram (def. 34), and because they are on equal bases BC and EF, and between the same parallels GH and BF, therefore the parallelogram GBCA= the parallelogram DEFH (I. 36).

Next, because the triangle ABC is half the parallelogram GBCA, and because the triangle DEF is half the parallelogram DEFH (I. 34),

Therefore, it is proved, as required, that (ax. 7)

The triangle ABC the triangle DEF.

Wherefore,

Triangles on equal bases, &c.

Q. E. D.

PROP. XXXIX. THEOREM.

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

Let ABC and DBC be equal triangles, on the same base BC and on the same (viz. the upper) side of it.

Then it is to be proved that

The triangles ABC and DBC are between the same parallels.

E

CONSTRUCTION.-1. Draw AD joining the vertices of the triangles (post. 1); then, if AD is not parallel to their base BC, 2. Through A draw AE parallel to this base BC, meeting BD in E (I. 31), and join EC.

PROOF.-Because the triangles ABC and EBC are on the same base BC (hyp.), and between the same supposed parallels AE and BC, therefore the triangle ABC = the triangle EBC (I. 37).

But the triangle ABC the triangle DBC (hyp.), therefore, also, the triangle DBC the triangle EBC (ax. 1), i.e. the whole its part, which is impossible (ax. 9).

=

Therefore the supposition that AE is parallel to BC is erroneous; and similarly it can be proved that only AD, joining the vertices of the triangles, is parallel to their base BC.

Therefore, it is proved, as required, that

The triangles ABC and DBC are between the same parallels.

Wherefore,

Equal triangles on the same base, &c.

Q. E. D.

PROP. XL. THEOREM.

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

Let ABC and DEF be equal triangles, on equal bases BC and EF, in the same straight line BF, and on the same (viz. the upper) side of it.

Then it is to be proved that

The triangles ABC and DEF are between the same parallels.

D

CONSTRUCTION.-1. Draw AD, joining the vertices of the triangles (post. 1); then, if AD is not parallel to the line of their bases BF,

2. Through A draw AG parallel to this line BF, meeting ED in G (I. 31), and join GF.

PROOF.-Because the triangles ABC and GEF are upon equal bases BC and EF (hyp.), and between the same supposed parallels AG and BF, therefore the triangle ABC= the triangle GEF (I. 38).

But the triangle ABC = the triangle DEF (hyp.), therefore, also, the triangle DEF the triangle GEF (ax. 1), i.e. the whole its part, which is impossible (ax. 9).

=

=

Therefore the supposition that AG is parallel to BF is erroneous; and similarly it can be proved that only AD, joining the vertices of the triangles, is parallel to the line of their bases, BF.

Therefore, it is proved, as required, that

The triangles ABC and DEF are between the same parallels.

Wherefore,

Equal triangles, on equal bases, &c.

Q. E. D.