problem-To constitute under the same conditions a rhomboid equivalent to the difference between given figures. The smaller rhomboid is here placed below the summit of the other, leaving the defect standing on the original base.

PROP. IX. THEOR.

A trapezoid is equivalent to the rectangle contained by its altitude and half the sum of its parallel sides.

The trapezoid ABCD is equivalent to the rectangle contained by its altitude and half the sum of the parallel sides BC and AD.

For draw CE parallel to AB (Í. 23.), bisect ED (I. 7.) in F; and draw FG parallel to AB, meeting the production of BC in G.

G

Because BC is equal to AE (I. 26.), BC and AD are. together equal to AE and AD, or to twice AE with ED, or to twice AE and twice EF, that is, to twice AF; consequently AF is half the sum of BC and AD. Wherefore the rectangle contained by the altitude of the trapezoid and half the sum of its parallel sides, is equivalent to the rhomboid BF (II. 1. cor.); but the rhomboid EG is equivalent to the triangle ECD (II. 5.), add to each the rhomboid BE, and the rhomboid BF is equivalent to the trapezoid ABCD.

E F D

Schol. Hence the area is found of any rectilineal figure referred to a given base; for it is equal to that of the aggregate rectangles under the mean of each pair of perpendiculars and the interjacent portion of the base.-This proposition is of great use in surveying, since it abridges the

mensuration of the irregular borders of a field, by help of what are called offsets, or perpendiculars branching from the great line to each remarkable flexure of the extreme boundary.

PROP. X. THEOR.

The square described on the hypotenuse of a right-angled triangle, is equivalent to the squares of the two sides.

Let the triangle ABC be right-angled at B; the square described on the hypotenuse AC is equivalent to BF and BI the squares of the sides AB and BC.

For produce DA to K, and through B draw MBL parallel to DA (I. 23.) and meeting FG produced in L.

F

B

I

Because the angle CAK, adjacent to CAD, is a right angle, it is equal to BAF: from each of these take away the angle BAK, and there remains the angle BAC equal to FAK. But the angle ABC is equal to AFK, both of them being right angles. Wherefore the triangles ABC and AFK, thus having two angles of the one respectively equal to those of the other, and the interjacent side AF equal to AB, are equal (1. 20.), and consequently the side AC is equal to AK. Hence the rectangle or rhomboid AM is equivalent to ABLK (II. 2. cor.), since they stand on equal bases AD and AK, and between the

M

E

same parallels DK and ML. But ABLK is (II. 1. cor.) equivalent to the rhomboid or square BF, for it stands on the same base AB and between the same parallels FL and AH. Wherefore the rectangle AM is equivalent to the square of AB.

And in like manner, by drawing MB to meet the production of HI, it may be proved, that the rectangle CM is equivalent to the square of BC. Consequently the whole square, ADEC, of the hypotenuse, contains the same space as both together of the squares described on the two sides AB and BC.

Cor. Hence the square of a side AB is equivalent to the rectangle under the hypotenuse AC and the adjacent segment AN made by a perpendicular.

Schol. This proposition is deservedly the most celebrated of the whole Elements, and serves as the main link for connecting Geometry with the modern Algebra.-The demonstration may be variously modified; but one of the simplest forms is that in which CAKO is proved to be a square, and the rectangle NK equivalent to the rhomboid AL and to the square BF on the one side, while the remaining rectangle NO is equivalent to the rhomboid CL and to the square BI on the other.

PROP. XI. THEOR.

If the square of one side of a triangle be equivalent to the squares of both the other sides, that side subtends a right angle,

Let the square described on AC be equivalent to the two squares of AB and BC; the triangle ABC is rightangled at B.

For draw BD perpendicular to AB (I. 34.) and equal to BC, and join AD.

Because BC is equal to BD, the square of BC is equal to the square of BD, and consequently the squares of AB and BC are equal to the squares of AB and BD. But the squares of AB and BC are, by hypothesis,

A

B

equivalent to the square of AC; and since ABD is, by construction, a right angle, the squares of AB and BD are, by the preceding proposition, equivalent to the square of AD. Whence the square of AC is equivalent to that of AD, and the straight line AC equal to AD. The two triangles ACB and ADB, having all the sides in the one respectively equal to those in the other, are therefore equal (I. 2.), and consequently the angle ABC is equal to the corresponding angle.ABD, that is, to a right angle.

Cor. Hence the numbers 3, 4, and 5 will express the sides and hypotenuse of a right-angled triangle-a property which readily suggests another method of erecting a perpendicular at the extremity of a straight line.

PROP. XII. PROB.

To find the side of a square equivalent to any number of given squares.

Let A, B, and C be the sides of the squares, to which it is required to find an equivalent square.

Draw DE equal to A, and from its extremity E erect (I. 34.) the perpendicular EF equal to B, join DF, and again, perpendicular to this, draw FG equal to C, and join DG: DG is the side of the square which was required.

For, since DEF is a right-angled triangle, the square of DF is equivalent to the squares of DE and EF (II. 10.), or of the lines A and B. Add on both sides the square of FG or of C, and the squares of DF and FG, which are equivalent to the square of DG (II. 10.), are equivalent to the aggregate squares of A, B, and C. And, by thus repeating the process, it may be extended to any number of squares.

PROP. XIII. PROB.

C

A
B

To find the side of a square equivalent to the difference between two given squares.

Let A and B be the sides of two squares; it is required to find a square equivalent to their difference.

Draw CD equal to the smaller line B, from its extremity erect (I. 34.) the indefinite per

A

pendicular DE, and about the cen

tre C, with a distance equal to the greater line A, describe a circle cutting DE in F: DF is the side of the square required.

For join CF. The triangle CDF being right-angled, the square of its hypotenuse CF is equivalent to

square

the squares of CD and DF (II. 10.), and consequently taking the square of CD from both, the excess of the of CF above that of CD is equivalent to the square of DF, or the square of DF is equivalent to the excess of the square of A above that of B.