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Parallelograms upon the same base and between the same parallels, are equal.

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Let the parallelograms ABCD and EBCF, be on the same base BC, and between the same parallels AF and BC; they are equal to one B another.

體述

L B

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For, produce the base BC to L; and because the line BL falls on the parallels AB and D C, the angles A B C and DCL are equal (Prop. 29.); also, because BL falls on the parallels E B and FC, the angles EBC and FCL are equal (Prop. 29.), therefore the angles ABE and DCF are also equal (Ax. 3.); and the lines which contain them are equal, AB to DC and EB to FC (Prop. 34.); therefore the triangles ABE and DCF are equal (Prop. 4.): take away each of these equal triangles respectively from the quadrilateral figure ABCF, and the remainders are the parallelograms ABCD and EBC F, which are equal (Ax. 3.). W. W. D.

SCHOL. On this proposition is founded the rule in Mensuration for finding the area of any parellelogram; namely, multiply the length of the base by the perpendicular breadth, and the product will be the area.

PROP. XXXVI.

THEOR.

Parallelograms upon equal bases and between the same parallels

are equal.

Let the parallelograms A B C D and EFGH be upon equal bases B C and FG, and between the same parallels AH and BG: they are equal.

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B

C

F

G

Join BE and CH; then the line EH is equal to FG (Prop. 34.), and BC is also equal to FG (Hyp.), therefore EH and BC are equal to one another but they are also parallel (Hyp.); therefore вE and CH are also equal and parallel (Prop. 33.); therefore вCHE is a parallelogram (Def. 35.); and it is equal to each of the parallelograms ABCD and EFGH, being on the same base and between the same parallels (Prop. 35.); therefore the parallelograms ABCD and EFGH are equal (Ax. 1.). W.W.D.

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Triangles upon the same base and between the same parallels are equal.

Let the triangles BAC and BDC be on the same base в C, and between the same parallels B C and AD: they are equal.

are

A

E P

D

B

Through the point c draw the line CE parallel to BA (Prop. 31.), and through B draw BF parallel to CD: then the parallelograms ABCE and FBCD equal, being on the same base BC, and between the same parallels (Prop. 35.); but the triangles B A C and BDC are halves of them (Prop. 34.), and are therefore equal. was to be demonstrated.

Which

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Triangles upon equal bases and between the same parallels are

equal.

Let the triangles B A C and DFE be upon equal bases B C and D E, and between the same parallels AG and BE; the triangles BAC and DFE are equal.

Through the point c draw the line CH parallel to AB (Prop. 31.), and through E draw E G parallel to DF; then the parallelo

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grams ABCH and FDEG are equal, being on equal bases and between the same parallels (Prop. 36.); but the triangles BAC and DFE are halves of them (Prop. 34.), and are therefore equal. Which was to be demonstrated.

COR. 1. A line drawn from the vertex of a triangle bisecting the base, bisects the triangle.

COR. 2. A triangle may be divided into any number of equal parts, by dividing the base into the given number of parts, and drawing right lines to the vertex of the triangle.

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

Let the equal triangles B A C and BDC be upon the same base BC, and on the same side of it: they are between the same parallels.

A

B

F

Ꭰ .

F

C

Join the points A and D; then, if AD be not parallel to BC, draw through a a line AF parallel to BC (Prop. 31.), cutting BD or BD produced in r: join cr; then, because the triangles BAC and BFC are upon the same base BC, and between the same parallels BC and AF, they are equal (Prop. 37.); but the triangle BAC is equal to BDC (Hyp.); therefore the triangles BFC and BDC are equal, a part equal to the whole, which is absurd: therefore AF is not parallel to BC; in the same manner it can be shown that no other line but AD is parallel to BC; therefore AD and BC are parallel. W. W. D.

PROP. XL. THEOR.

Equal triangles upon equal bases, in the same right line, and on the same side, are between the same parallels.

Let the equal triangles BAC and EDG be upon equal bases BC and EG, in the same line BG, and on the same side of it: they are between the same parallels.

B

F

C E

Join the points A and D; then if AD be not parallel to BG, draw through A a line AF parallel to BG (Prop. 31.), cutting ED or ED produced in F: join GF; then, because the triangles BAC and EFG are upon equal bases BC and EG, and between the same parallels BG and AF, they are equal, (Prop. 38.); but the triangle BAC is equal to EDG (Hyp.); therefore the triangles EFG and EDG are equal, a part equal to the whole, which is absurd; therefore AF is not parallel to BG; in the same manner it can be shown that no other line but AD is parallel to BC; therefore A D and BG are parallel. W. W. D.

C

PROP. XLI. THEOR.

A parallelogram is double of a triangle upon the same base and between the same parallels.

Let the parallelogram A B C D and the triangle BEC be upon the same base BC, and between the same parallels AE and BC: the parallelogram is double of the triangle.

A

D

E

B

C

Join AC: then, because the triangles BAC and BEC are upon the same base BC, and between the same parallels AE and BC, they are equal (Prop. 37.); but the parallelogram ABCD is double of the triangle BAC (Prop. 34.); therefore the parallelogram ABCD is also double of the triangle BEC. Which was to be demonstrated. SCHOL. Hence the rule in Mensuration for finding the area of a triangle: multiply the length of the base by the perpendicular breadth of the triangle, and half the product will be the area.

PROP. XLII. PROB.

To describe a parallelogram equal to a given triangle, and having an angle equal to a given one.

Let BAC be the given triangle, and D the given angle. It is required to describe a parallelogram equal to the triangle BA C, and having an angle equal to D.

A

B

G F

E C

Draw through A a line AF parallel to BC: bisect the base BC in E, and with the line EC make an angle cEG equal to the given angle D (Prop. 23.): from c draw CF parallel to EG, and join AE: then, because GF is parallel to EC, and CF parallel to EG (Const.), ECFG is a parallelogram (Def. 35.): and because the triangles BAE and EAC are upon equal bases BE and EC, and between the same parallels, they are equal (Prop. 38.); therefore the whole triangle BAC is double of one of them E AC: but the parallelogram ECFG is also double of the triangle EAC (Prop. 41.); therefore the parallelogram ECFG is equal to the triangle BAC (Ax. 6.); and it has an angle CEG equal to the given angle D (Const.) W. W. D.

PROP. XLIII. THEOR.

In a parallelogram the complements of the parallelograms about the diagonal are equal.

Let ABCD be a parallelogram, and AC the diagonal; also let FE and HL be the parallelograms about the diagonal; then the complements BG and GD which complete the figure are equal.

F.

B H

G

E

D

L

Because each parallelogram is divided by its diagonal into two equal parts (Prop. 34.), the triangle CAB is equal to CAD, the triangle GAF to GAE, and the triangle CGH to CGL; therefore the two triangles GAE and CGL are equal to the two triangles GAF and CGH; take away these equals from the equal triangles CAD and CA B, and the remainders are the complements BG and GD, which are equal (Ax. 3.). W. W. D.

PROP. XLIV. PROB.

To a given right line to apply a parallelogram, equal to a given triangle, and having an angle equal to a given one.

Let ABC be the given triangle, cr the given line, and D the given angle; it is required to apply a parallelogram to the line CF, equal to the triangle ABC, and having an angle equal to the angle D.

B

A G

D

H

F

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N

P

Place one side BC of the triangle in the same line with CF so that they form a continued right line; bisect BC in E, and make the parallelogram CEGH equal to the triangle ABC, and having an angle ECH equal to the angle D (Prop. 42.); through F draw FL parallel to CH or EG, and produce GH to L: draw LC to meet GE produced to N; through N draw NP parallel to EF or GL; produce LF and HC to P and s.

Then in the parallelogram GP, the complements GC and CP are equal (Prop. 43.); but Gc is equal to the triangle ABC (Const.), therefore CP is equal to the triangle ABC.

And because the angle ECH is equal to the angle FCS (Prop. 15.), and also to the angle D, therefore the angle rcs is equal to the angle D (Ax. 1.).

Therefore the parallelogram CP is applied to the line cr, is equal to the triangle ABC, and has an angle rcs equal to the given angle D. W. W. D.

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