equal (I. 24.). To each of them, add the quadrilateral space ABCE, and the rectangle AFCE is equal to the tra pezium ABCD.

PROP. XIII. THEOR.

A trapezoid is equivalent to the rectangle contained by its altitude and half the sum of its parallél 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 (I. 26.), bisect ED (I.7.) in F, and draw FG parallel to AB, meeting the production of BC in G.

B

G

Because BC is equal to AE (I. 29.), BC and AD are together equal to AE and AD, or to twice AE and 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

E F

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. 7.), add to each the rhomboid BE, and the rhomboid BF is equivalent to the trapezoid ABCD.

Cor. Hence the greater of two lines is equal to half their sum and half their difference; for AD is equal to AF joined to FD, which is half the difference ED. The smaller line AE again is formed by taking half the difference from half the sum.

PROP. XIV. THEOR.

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

the two sides.

squares of

Let ACB be a triangle which is right-angled at B; the square of the hypotenuse AC is equivalent to the two squares of AB and BC.

For produce the base BA until AD be equal to the perpendicular BC, and on DB describe (I. 39.) the square DEFB, make GE and FH equal to AD or BC, join AG, GH, and HC, and through the points A and C (I. 26.) draw AL and CI parallel to BF and BD.

E

LH

F

Because the whole line BD is equal to DE, and a part of it AD equal to GE, the remainder AB is equal to DG; wherefore the triangles ACB and AGD are equal (I. 3.), since they have the sides AB, BC equal to DG, AD, and the contained angle ABC equal to ADG, both of these being right angles. In the same manner, it is proved, that the triangle ACB is equal to GEH, and to HFC. Consequently the sides AC, AG, GH, and HC are all equal. But G the angle CAB, being equal to AGD, is equal to the alternate angle GAL (I. 29.); add LAC to each, and the whole angle LAB or EDB (I. 29.) is equal to GAC, which is, therefore, a right angle. Hence the figure AGHC, having all its sides equal and one angle right, is a square.

I

D

C

K

B

Again, the parallelograms KB and KE are evidently rectangular; they are also equal, being contained by equal sides; and each of them being double of the original tri

angle ACB, they are together equal to the four triangles ACB, AGD, EHG, and HCF. The other inscribed figures LC and IA are obviously the squares of KC and AD, which are equal to the base and perpendicular of the triangle ABC. From the whole square DEFB, therefore, take away separately those four encompassing triangles, and the two interjacent rectangles KB and KE, and the remainders must be equal; that is the square AGHC is equal in space to both the squares ADIK and KLFC.

Otherwise thus.

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. 26.) and meeting FG produced in L.

Because the angle CAK, adjacent to CAD, is a right angle, it is equal to BAF; from each take away the angle BAK, and there remains the

angle BAC equal to FAK. But the angle ABC is equal to AFK, both being right angles. Wherefore the triangles ABC and AFK, having thus two angles of the one respectively equal to those of the other, and the interjacent side AF equal to AB, are equal (I. 23.), 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 same parallels

DK and ML. But ABLK is equivalent to the square or rhomboid BF (II. 1. cor.), 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.

PROP. XV. THEOR.

If the square of a side of a triangle be equivalent to the squares of both the other sides, that side is the hypotenuse of a right-angled triangle.

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. 38) and equal to BC, and join AD.

B

Because BC is equal to BD, the square of BC is (II. 9. cor.) 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 equivalent to the square of AC; and since ABD is, by construction, a right angle, the squares of AB and BD are (II. 12.) equivalent to the square of AD. Whence the square of AC is equivalent to that of AD, and (II. 9. cor.) the straight line AC equal to CD. 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.

PROP. XVI. 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. 38.) the perpendicular EF equal to B, join DF, and perpendicular to this draw FG equal to C, and join DG: DG is the side of the square which was requi

red.

D

T

For because DEF is a right-angled triangle, the square of DF is equivalent to the squares of DE and EF (II. 14.), or of A and B. Add the square of FG or C, and the squares of DF and FG, which are equivalent to the square of DG (II. 14.), 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.

C

A

E

B

PROP. XVII. PROB.

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.