PROP. XLI. THEOREM.

D

E

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 BD, and the triangle EBC A be upon the same base BC, and between the same parallels BC, A E. The parallelogram BD is double of the triangle EBC.

B

Join A C. The triangle A B C is equal to the triangle EBC (I. 37), because they are upon the same base BC, and between the same parallels BC, AE. But the parallelogram BD is double of the triangle ABC (I. 34), because the diagonal A C bisects it. Therefore the parallelogram B D is also double of the triangle EBC. Therefore, if a parallelogram and a triangle, &c. Q. E. D.

This proposition is the foundation of the mensuration of triangles, and consequently of all rectilineal figures, as they can easily be divided into triangles. Inasmuch as the area of a parallelogram is found by multiplying its base by its perpendicular altitude or breadth, so the area of a triangle is found by multiplying its base by its perpendicular altitude, and taking half the product. In a triangle the perpendicular altitude is the perpendicular drawn from the vertex of the angle opposite to the base, to the base itself or to the base produced.

Corollary.-A parallelogram is equal to a triangle on the double of its base, and between the same parallels.

Exercise 1.-If from any point within a parallelogram, straight lines be drawn to the extremities of two opposite sides, the two triangles upon these sides are together equal to half of the parallelogram.

Exercise 2.-In a trapezoid, if one of the sides which is not parallel be bisected and straight lines be drawn from the point of bisection to the extremities of the other side which is not parallel, the triangle which they form with the latter side is equal to half the trapezoid.

To describe a parallelogram equal to a given triangle, and having 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 equal to the given triangle A B C, and having one of its angles equal to D.

Bisect BC in E (I. 10), and join A E. At the point E in the straight line EC, make the angle CEF (I. 23) equal to the angle D. Through A draw AFG (I. 31), parallel to B C, and through C draw CG parallel to E F. Then the figure CEFG is a parallelogram (Def. 36).

B

A F G

Ε C

Because the two triangles A BE, AEC are on equal bases BE, E C, and between the same parallels B C, A G; they are equal (I. 38) to one another. Therefore the triangle ABC is double of the triangle AE C. But the parallelogram FC is double of the triangle AEC (I. 41), because they are upon the same base EC, and between the saine parallels E C, A G. Therefore the parallelogram

FC is equal (Ax. 6) to the triangle ABC, and it has one of its angles CEF equal to the given angle D. Wherefore, a parallelogram FC has been described equal to the given triangle ABC, and having one of its angles CEF equal to the given angle Ď. Q. E. D. Exercise. To describe a triangle equal to a given parallelogram, and having an angle equal to a given rectilineal angle.

PROP. XLIII. THEOREM.

The complements of the parallelograms, which are about the diagonal of any parallelogram, are equal.

Let BD be a parallelogram, of which the diagonal is A C; and EH, GF the parallelograms about A C, that is, through which AC passes. Then BK, KD are the other parallelograms which make up the whole figure BD, and are therefore called the complements. The complement BK is equal to the complement K D.

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K

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F

Because BD is a parallelogram, and A C its diagonal, the triangle ABC is equal (I. 34) to the triangle AD Č. Again, because ÈH is a parallelogram, and A Kits diagonal, the triangle AEK is equal (I. 34) to the triangle AH K. For the same reason, the triangle KGC is equal to the triangle KFC. Therefore the two triangles A EK, KGC are equal (Ax. 2) to the two triangles AHK, KFC. But the whole triangle ABC is equal to the whole triangle ADC. Therefore the remaining complement BK is equal (Ax. 3) to the remaining complement K D. Wherefore the complements, &c. Q. E. D.

Corollary 1.-The parallelograms about the diagonal of a parallelogram, as also its complements, are equiangular to the whole parallelogram and to each other.

Corollary 2.-If through any point in the diagonal of a parallelogram, straight lines be drawn parallel to its sides, the parts into which each divides the parallelogram are equal, the greater to the greater, and the less to the less. Exercise 1.-In the preceding figure join EH, BD, and GF, and demonstrate that the three diagonals thus drawn, are parallel to each other.

Exercise 2.-If about the diagonal of a parallelogram, any number of parallelograms be placed, whether the extremities of their diagonals coincide in the same point or not, the remaining complementary rectilineal figures, on each side of the diagonal of the parallelogram, are equal.

To a given straight line to apply a parallelogram, equal to a given triangle, and having one of its angles equal to a given rectilineal angle.

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

Make the parallelogram BF equal (I. 42) to the triangle C, and

H

A

Because the straight line HF falls upon the two parallels AH, EF, the two angles AHF, HFE are together equal (I. 29) to two right angles. Therefore the two angles BHF, HFE are less than two right angles. But those straight lines which with another straight line, make the two interior angles upon the same side of it less than two right angles, meet (Ax. 12) if produced far enough. Therefore HB, FE shall meet, if produced. Let them be produced and meet in K. Through K draw KL parallel to EA or FH, and produce HA, GB, to meet KL in the points L and M. Because FL is a parallelogram (Def. 36), of which the diagonal is HK; AG, ME, are the parallelograms about HK, and LB, BF are the complements. Therefore the complement LB is equal (I. 43) to the complement BF. But the complement BF is equal (Const.) to the triangle C. Therefore LB is equal (Ax. 1) to the triangle C. Because the angle GBE is equal (L. 15) to the angle ABM, and likewise (Const.) to the angle D. Therefore the angle ABM is equal (Ax. 1) to the angle D. Wherefore to the straight line AB, is applied the parallelogram L B, equal to the triangle C, and having the angle ABM equal to the angle D. Q. E. F.

Corollary. From this proposition, it is manifest how to describe on a given straight line, a rectangle equal to a given triangle.

Exercise. To a given straight line to apply a triangle equal to a given parallelogram and having an angle equal to a given rectilineal angle.

To describe a parallelogram equal to a given rectilineal figure and having an angle equal to a given rectilineal angle.

Let AB CD be the given rectilineal figure, and E the given rectilineal angle. It is required to describe a parallelogram equal to the figure ABCD, and having an angle equal to E.

Join DB. Describe the parallelogram A FH equal to the triangle ADB, and having the angle FKH equal (I. 42) to the angle E. To the straight line GH apply the parallelogram GM equal to the triangle DB C, and having the angle GHM equal (I. 44) to the angle E. Then the figure KL is the parallelogram required.

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F G L

E

Because the angle E is equal (Const.) to each of the angles FKH, GHM, the angle FKH is equal (Ax. 1) to the angle GHM. To each of these equals, add the angle KHG. Therefore the two angles FKH, KHG are equal to the two angles KHG, GHM. But the two angles FKH, KHG are equal (I. 29) to two right angles. Therefore the two angles

KHG, GHM are also equal (Ax. 1) to two right angles. Because at the point H, in the straight line GH, the two straight lines KH, HM, upon opposite sides of it, make the adjacent angles equal to two right angles HK is in the same straight line (I. 14), with HM. Again, because the straight line HG meets the parallels KM, FG, therefore the angle MHG is equal (I. 29) to the alternate angle HGF. To cach of these equals, add the angle HGL. Therefore the two angles MHG, HGL are equal to the two angles HGF, HGL; but the two angles MHG, HGL are equal (I.29) to two right angles. Therefore also the two angles HGF, HGL are equal (Ac. 1) to two right angles. Wherefore, as before, FG (I. 14) is in the same straight line with GL. Because KF is parallel to HG, and HG to ML, KF is parallel (I. 30) to ML. And KM has been proved parallel to FL. Therefore the figure KL is a parallelogram (Def. 36). But the parallelogram FH, is equal (Const.) to the triangle ABD, and the parallelogram GM to the triangle BDC. Therefore the whole parallelogram KL, is equal to the whole rectilineal figure ABCD. Wherefore the parallelogram KL has been described equal to the given rectilineal figure ABCD, and having the angle F K M equal to the given angle E. Q. E. F.

Corollary. From this it is manifest how, to a given straight line, to apply a parallelogram, which shall have an angle equal to a given rectilineal angle, and shall be equal to a given rectilineal figure; viz., by applying to the given straight line a parallelogram equal to the first triangle ABD (I. 44), and having an angle equal to the given angle, and constructing the rest of the figure as in this proposition.

By this proposition, also, upon a given straight line, a rectangle may be described, equal to a given rectilineal figure.

To describe a square upon a given straight line.

Let AB be the given straight line. to describe a square upon AB.

It is required

From the point A draw A C at right angles (I. 11) to AB. Make A D equal (I. 3) to A B. Through the point D draw DE parallel (I. 31) to A B, and through the point B draw B E parallel to AD. The figure AE is a square.

Because AE is a parallelogram (Def. 36) A B is equal (I. 34) to DE, and AD to BE. But BA is equal to AD. Therefore the four lines BA, AD,

DE, EB are all equal to one another, and the parallelogram AE is equilateral. Again, because AD meets the parallels A B, DE, the two angles BAD, ADE, are equal (I. 29) to two right angles. But BAD is (Const.) a right angle. Therefore also A D E is a right angle. But the opposite angles of parallelograms (I. 34) are equal. Therefore each of the opposite angles ABE, BED, is a right angle, and the parallelogram AE is rectangular, that is, has all its angles right angles. And it has been proved to be equilateral. Therefore the figure A E is a square (Def. 30), and it is described upon the given straight line AB. Q. E. F.

Corollary 1.-Hence, every parallelogram that has one right angle, has all its right angles.

The construction of this problem may be effected, in accordance with Euclid's definition of a square, by Prop. XI. of this book. Euclid's corollary has also been anticipated.

Corollary 2.-If two squares be equal, their sides, or the straight lines on which they are described, are equal.

Exercise.-If, in the sides of a square, points be taken at equal distances from its four angular points, in succession, the straight lines which join these points in the same order, will form a square. This square will be less than the original square in proportion to the distance of the four assumed from its angular points, until that distance be equal to half the side of the square, when the inscribed square (Def. 1, Book IV.) will be a minimum,—that is, the least possible square which can be thus inscribed.

In any right-angled triangle, the square described upon the side subtending the right angle, is equal to the squares described upon the two sides, containing the right angle.

Let ABC be a right-angled triangle, having the right angle B A C, The square described upon the side BC is equal to the squares described upon the two sides BA and A C.

Upon the side B C describe the square BE (I. 46), and on the sides BA and A C, the squares GB, and H C. Through A draw AL parallel to BD or F CE (I. 31) and join AD and FC.

B

G

D LE

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Эк

C

Because the angle BAC is a right angle (Hyp.), and that the angle BAG is a right angle (Def. 30), the two straight lines AC, AG, upon the opposite sides of AB, make with it, at the point A, the adjacent angles equal to two right angles. Therefore CA is (I. 14) in the same straight line with AG. For the same reason, B A and AH are in the same straight line. Because the angle DBC is equal to the angle FB A, each of them being a right angle. To each of these equals, add the angle A B C. Therefore the whole angle DBA is equal (Ax. 2) to the whole angle FBC. And because the two sides AB, BD, are equal to the two sides FB, BC, each to each, and the angle ABD to the angle FBC. Therefore the base AD is equal to the base FC (I. 4), and the triangle ABD to the triangle F B C. But the parallelogram B L is double of the triangle ABD (I. 41), because they are upon the same base BD, and between the same parallels BD, and AL. Also the square GB is double of the triangle FBC, because these are upon the same base FB, and between the same parallels FB and G C. Now, the doubles of equals are equal to one another (Ax. 6). Therefore the parallelogram BL is equal to the square GB. Similarly, by joining A E and BK, it can be proved that the parallelogram CL is equal to the square HC. Therefore the whole square BE is equal (4x. 2) to the two squares G B and H C. Wherefore the square described upon the side B C opposite to the right angle BA C, is equal to the squares described upon the two sides A B and A C, containing Therefore, in any right-angled triangle, &c. Q. E. D.