PROP. XXXV. THEOR.

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

Let the parallelograms ABCD, EBCF (figs. 2. and 3.) be upon the same base BC, and between the same parallels AF, BC; these parallelograms are equal to one another.

D

F

If the sides AD, DF, of the parallelograms AC, BF (fig. 1.) opposite to the base BC, be terminated in the same point D; each of the parallelograms is double (I. 34.) of the triangle BDC; and they are therefore equal (I. ax. 6.) to one another.

B

C

But if the sides AD, EF (figs. 2. and 3.) opposite to the base BC, be not terminated

A

DE

C

F A

E D

P

B

C

in the same point; then, because AC is a parallelogram, DC is equal (I. 34.) to AB; also (I. 29. part 2.) because AB is parallel to DC, the angles FDC, EAB are equal; and because BE is parallel to CF, the angles DFC, AEB are likewise equal; wherefore (I. 26. part 3.) the triangles FDC, EAB are equal to one another. Take the triangle FDC from the trapezium ABCF, and from the same trapezium, take the triangle EAB, and (I. ax. 3.) the remainders are equal; that is, the parallelogram AC is equal to the parallelogram BF.* Therefore parallelograms, &c.

PROP. XXXVI. THEOR.

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

Let AC, EG be parallelograms upon equal bases BC, FG, and between the same parallels AH, BG; the parallelogram AC is equal to EG.

A

DE

B

C F

H

G

Join BE, CH; and because BC is equal (hyp.) to FG, and (I. 34.) FG to EH; BC is equal to EH; and (hyp.) they are parallel, and they are joined towards the same parts by the straight lines BE, CH; therefore (I. 33.) BE, CH are both equal and parallel, and (I. def. 24.) EBCH is a parallelogram; and it is equal (I. 35.) to AC, because

*It will appear, from this proposition, that the perimeters of two equal parallelograms on the same base may differ in any degree whatever; and (I. 19.) that, of all equal parallelograms on the same base, the rectangle has the least perimeter.

It may also be remarked that Legendre and some other writers call only such figures equal, as exactly coincide when applied to one another; while those which, as in this proposition, "without coinciding, contain the same space," they call equivalent. This seems to be an unnecessary refinement.

it is upon the same base BC, and between the same parallels BC, AH. For the like reason, the parallelogram EG is equal to the same parallelogram EBCH; because it is on the same base EH, and between the same parallels EH, BG. Therefore also the parallelogram AC is equal (I. ax. 1.) to EG. Wherefore parallelograms, &c.

PROP. XXXVII. THEOR.*

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

B

A D

F

Let the triangles ABC, DBC be upon the same base BC, and between the same parallels AD, BC; ABC is equal to DBC. Produce AD both ways to E, F, and through B draw (I. 31.) BE parallel to CA; and through C,+ draw CF parallel to BD. Therefore (I. def. 24.) each of the figures EC, BF is a parallelogram; and EC is equal (I. 35.) to BF, because they are upon the same base BC, and between the same parallels BC, EF. But the triangle ABC is the half of the parallelogram EC, because (I. 34.) the diagonal AB bisects it; and the triangle DBC is the half of the parallelogram BF, because the diagonal DC bisects it: but (I. ax. 7.) the halves of equals are equal; therefore the triangle ABC is equal to the triangle DBC. Wherefore triangles, &c.

PROP. XXXVIII. THEOR.

TRIANGLES 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: ABC is equal to DEF.

G

A

D

H

Produce AD both ways to the points G, H, and (I. 31.) through B draw BG parallel to CA, and through F draw FH parallel to ED. Then (I. def. 24.) each of the figures GC, EH, is à parallelogram; and (I. 36.) they are equal to one another, because they are upon equal bases BC, EF, and between the same parallels BF, GH; and (I. 34.) the triangle ABC is half of the parallelogram GC; and

B

CE

F

* This proposition and the following afford instances in which triangles, though not, as in propositions 4, 8, 26, in every respect equal, are equal in magnitude.

That BE and DA will meet when produced, follows from the 12th axiom, the angles EAB, EBA being equal (I. 29.) to two angles of the triangle ABC, and consequently less than two right angles. In like manner it might be shown that AD and CF will meet. The like reasoning is applicable in the next proposition.

the triangle DEF of the parallelogram EH. But (I. ax. 7.) the halves of equals are equal; therefore the triangle ABC is equal to the triangle DEF. Wherefore triangles, &c.

Cor. 1. From this proposition and from the 36th, it is plain, that triangles and parallelograms between the same parallels, but upon unequal bases, are unequal; that which has the greater base being greater than the other.

Cor. 2. Hence a straight line drawn from the vertex of a triangle to the point of bisection of the base, bisects the triangle: and if two triangles have two sides of the one respectively equal to two sides of the other, and the contained angles (I. def. 38.) supplemental, the triangles are equal.

EQUAL triangles upon 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 are between the same parallels. Join AD; AD is parallel to BC. For, if it be not, through A draw (I. 31.) AE parallel to BC, and join EC. Then (I. 37.) the triangles ABC, EBC are equal, because they are on the same base BC, and between the same parallels BC, AE. But the triangle ABC is equal (hyp.) to the triangle DBC: therefore also the triangle EBC is equal to the triangle DBC, the less to the greater, which is impossible. Therefore AE is not parallel to BC; and in the same manner it can be demonstrated that no other line drawn through A, except AD, is parallel to BC; AD is therefore parallel to it. Wherefore equal triangles, &c.

PROP. XL. THEOR.

B

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

A

D

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

G

B

CE

F

Join AD: AD is parallel to BC. For, if it be not, through A draw (I. 31.) AG parallel to BF, and join GF. Then (I. 38.) the triangles ABC, GEF are equal, because they are on equal bases BC, EF, and between

In this proposition and the following, we have two other principles by which lines are proved to be parallel, in addition to those contained in propositions 27, 28, 30, and 33.

the same parallels BF, AG; but the triangle ABC is equal (hyp.) to the triangle DEF; therefore also the triangle GEF is equal to the triangle DEF, the less to the greater, which is impossible. Therefore AG is not parallel to BF; and in the same manner it can be demonstrated, that through A no other straight line can be drawn parallel to BF, but AD: AD is therefore parallel to BF. Wherefore equal triangles, &c.

PROP. XLI. THEOR.

If a parallelogram and a triangle be upon the same base, and between the same parallels; the parallelogram is double of the triangle.

Let the parallelogram ABCD and the triangle EBC be upon the same base BC, and between the same parallels BC, AE; the parallelogram is double of the triangle.

A

B

D E

Join AC. Then (I. 37.) the triangles ABC, EBC are equal, because they are upon the same base BC, and between the same parallels BC, AE. But (I. 34.) the parallelogram ABCD is double of the triangle ABC, because the diagonal AC bisects it: wherefore ABCD is also double of the triangle EBC. Therefore, if a parallelogram, &c.

PROP. XLII. PROB.

To describe a parallelogram equal to a given triangle, and having one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle. It is required to describe a parallelogram equal to ABC, and having one of its angles equal to D.

F

G

Bisect (I. 10.) BC in E, join AE, and (I. 23.) at the point E, in the straight line BC, make the angle CEF equal to D; through A draw (I. 31.) AG parallel to BC, and through C draw CG parallel to EF; FECG is the parallelogram required.

B E C D

For, since (const.) BE is equal to EC, the triangle ABE is equal (I. 38. cor. 2.) to the triangle AEC; therefore the triangle ABC is double of the triangle AEC. The parallelogram EG is likewise double (I. 41.) of the triangle AEC, because it is upon the same base, and between the same parallels. Therefore the parallelogram EG is equal (ax. 6.) to the triangle ABC, and it has (const.) one of its angles CEF equal to D; wherefore there has been described a parallelogram EG equal to a given triangle ABC, and having one of its angles CEF equal to the given angle D which was to be done.

PROP. XLIII. THEOR.

IF through any point in either diagonal of a parallelogram, straight lines be drawn parallel to the sides; of the four parallelograms thus formed, those through which the diagonal does not pass, and which are called the complements of the other two, are equal.

*

E

D

A H

F

K

B G

C

Let BD be a parallelogram, of which AC is one of the diagonals: then, if through K a point in AC, the two straight lines EF and HG be drawn parallel to the sides, the parallelograms BK and KD are equal. Because BD is a parallelogram, and AC its diagonal, the triangle ABC is equal (I. 34.) to the triangle ADC. And, because EH is a parallelogram, the diagonal of which is AK, the triangles AEK, AHK are equal. For the same reason, the triangles KGC, KFC are equal. Then, because the triangle AEK is equal to AHK, and KGC to KFC, the triangles AEK, KGC are together equal (I. ax. 2.) to the triangles AHK, KFC taken together. But the whole triangle ABC is equal to the whole ADC; therefore (I. ax. 3.) the remaining complements BK, KD are equal. † Wherefore, if through any point, &c.

PROP. XLIV. PROB.

To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given angle.

Let AB be the given straight line, C the given triangle, and D the given angle. It is required to apply to AB a parallelogram equal to C, and having an angle equal to D.

Make (I. 42.) the parallelogram BF equal to the triangle C, having the angle EBG equal to the angle D, and so that BE may

EH and GF are generally called the parallelograms about the diagonal, and BK and KD, the complements of EH and GF, because with these, they complete the entire parallelogram BD.

By adding first EH, and then GF, to the complements, we have the parallelograms ED and BF respectively equal to BH and GD: and hence it follows, that if lines, EF, HG, be drawn parallel to the sides, through any point K in the diagonal, the parts into which one of them divides the parallelogram, are respectively equal to those into which it is divided by the other.

To apply a parallelogram to a given straight line, is to describe a parallelogram having that line for one of its sides.