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equal parts, each less than EG; there will

FI с be at least one point of division between E and G. Let be that point, and draw the perpendicular HI. The bases AB, AH will be to each other in the ratio of two whole numbers, and by the preceding case

Α.

EHN B we shall have

ABCD: AHID :: AB: AH. But, by hypothesis, we have

ABCD: AEFD:: AB : AG. In these two proportions the antecedents are equal; therefore the consequents are proportional (Prop. IV., Cor., B. II.), and we have

AHID : AEFD :: AH : AG. But AG is greater than AH; therefore the rectangle AEFD is greater than AHID (Def. 2, B. II.) ; that is, a part is greater than the whole, which is absurd. Therefore ABCD can not be to AEFD as AB to a line greater than AE.

In the same manner, it may be shown that the fourth term of the proportion can not be less than AE; hence it must be AE, and we have the proportion

ABCD : AEFD :: AB : AE. Therefore, two rectangles, &c.

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Any two rectangles are to each other as the products of their bases by their altitudes.

Let ABCD, AEGF be two rectangles; the ratio of the rectangle ABCD to the rectangle AEGF, is the same with the ratio of the product of AB by AD, to th> product of AE by AF; that is,

ABCD: AEGF:: AB XAD: AEX AF. Having placed the two rectangles so that the angles at A are vertical, pro

H D duce the sides GE, CD till they meet in H. The two rectangles ABCD, AEHD have the same altitude AD; they are,

B therefore, as their bases AB, AE (Prop. III.). So, also, the rectangles AEHD,

I AEGF, having the same altitude AE, are to each other as their bases AD, AF Thus, we have the two proportions

ABCD : AEHD:: AB AE,
AEHD: AFGF :: AD AF.

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Hence (Prop. XI., Cor., B. II.),

ABCD: AEGF:: ABX AD: AEX AF. Scholium. Hence we may take as the measure of a rectangle the product of its base by its altitude; provided we understand by it the product of two numbers, one of which is the number of linear units contained in the base, and the other the number of linear units contained in the altitude.

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The area of a parallelogram is equal to the product of its base by its altitude.

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Let ABCD be a parallelogram, AF its

Е С altitude, and AB its base; then is its surface measured by the product of AB by

For, upon the base AB, construct a rectangle having the altitude AF; the par

Α.

B allelogram ABCD is equivalent to the rectangle ABEF (Prop. I., Cor.). But the rectangle ABEF is measured by ABX AF (Prop. IV., Schol.); therefore the area of the parallelogram ABCD is equal to AB XAF.

Cor. Parallelograms of the same base are to each other as their altitudes, and parallelograms of the same altitude are to each other as their bases; for magnitudes have the same ratio that their equimultiples have (Prop. VIII., B. II.).

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The area of a triangle is equal to half the product of its base by its altitude.

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Let ABC be any triangle, BC its base, and AD its altitude ; the area of the triangle ABC is measured by half the product of BC by AD.

For, complete the parallelogram ABCE. The triangle ABC is half of the parallelo

D C gram ABČE (Prop. II.) ; but the area of the B parallelogram is equal to BC XAD (Prop. V.); hence the area of the triangle is equal to one half of the product of BC by AD. Therefore, the area of a triangle, &c.

Cor. 1. Triangles of the same altitude are to each other as their bases, and triangles of the same base are to each other as their altitudes.

Cor. 2 Equivalent triangles, whose hases are equal, have equal altitudes; and equivalent triangles, whose altitudes are equal, have equal bases.

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The area of a trapezoid is equal to half the product of its ultitude by the sum of its parallel sides.

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Let ABCD be a trapezoid, DE its al

С titude, AB and CD its parallel sides; 'ts area is measured by half the product of DE, by the sum of its sides AB, CD.

Bisect BC in F, and through F draw GH parallel to AD, and produce DC to Á E G H. In the two triangles BFG, CFH, the side BF is equal to CF by construction, the vertical angles BFG, CFH are equal (Prop. V., B. I.), and the angle FCH is equal to the alternate angle FBG, because CH and BG are parallel (Prop. XXIII., B. I.); therefore the triangle CFH is equal to the triangle BFG. Now, if from the whole figure, ABFHD, we take away the triangle CFH, there will remain the trapezoid ABCD; and if from the same figure, ABFHD, we take away the equal triangle BFG, there will remain the parallelogram AGHD. Therefore the trapezoid ABCD is equivalent to the parallelogram AGHD, and is measured by the product of AĠ by DE.

Also, because AG is equal to DH, and BG to CH, therefore the sum of AB and CD is equal to the sum of AG and DH, or twice AG. Hence AG is equal to half the sum of the parallel sides AB, CD; therefore the area of the trapezoid ABCD is equal to half the product of the altitude DE by the sum of the bases AB, CD.

Cor. If through the point F, the middle of BC, we draw FK parallel to the base AB, the point K will also be the middle of AD. For the figure AKFG is a parallelogram, as also DKFH, the opposite sides being parallel. Therefore AK is equal to FG, and DK to HF. But FG is equal to FH, since the triangles BFG, CFH are equal; therefore AK is equal to DK.

Now, since KF is equal to AG, the area of the trapezoid is equal to DEXKF. Hence the area of a trapezoid is equal to its altitude, multiplied by the line which joins the middle points of the sides which are not parallel.

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If a straight line is divided into any two parts, the square of the whole line is equivalent to the squares of the two parts, together with twice the rectangle contained by the parts.

Let the straight line AB be divided into any two parts in C; the square on AB is equivalent to the squares on AC CB, together with twice the rectangle contained by AC, CB; that is,

AB', or (AC+CB)'=AC +CB'+2AC X CB. Upon AB describe the square ABDE;

H. D take AF equal to AC, through F draw FG parallel to AB, and through Č draw CH par- F allel to AE.

The square ABDE is divided into four parts: the first, ACIF, is the square on AC, since AF was taken equal to AC. The sec

С B ond part, IGDH, is the square on CB; for, because AB is equal to AE, and AC to AF, therefore BC is equal to EF (Axiom 3, B. I.). But, because BCIG is a parallelogram, GI is equal to BC; and because DEFG is a parallelogram, DG is equal to EF (Prop. XXIX., B. I.); therefore HIĞD is equal to a square described on BC. If these two parts are taken from the entire square, there will remain the two rectangles BCIG, EFIH, each of which is measured by ACX CB; therefore the whole square on AB is equivalent to the squares on AC and CB, together with twice the rectangle of AC XCB. Therefore, if a straight line, &c.

Cor. The square of any line is equivalent to four times the square of half that line. For, if AC is equal to CB, the four figures AI, CG, FH, ID become equal squares.

Scholium. This proposition is expressed algebraically thus :

(a+b)=a2+2ab+b'.

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The square described on the difference of two lines, is equivalent to the sum of the squares of the lines, diminished by twice the rectangle contained by the lines.

Let AB, BC be any two lines, and AC their difference: the square described on AC is equivalent to the sum of the squares on AB and CB, diminished by twice the reciangle contained by AB, CB; that is,

AC", or (AB-BC)'=AB'+BC-2AB X BC.
Upon AB describe the square ABKF; L F
take AE equal to AC, through C draw
CG parallel to BK, and through E draw

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D HI parallel to AB, and complete the I E

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square EFLI.

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Because AB is equal to AF, and AC to AE; therefore CB is equal to EF, and GK

Св to LF. Therefore LG is equal to FK or AB; and hence the two rectangles CBKG, GLID are each measured by ABX BC. If these rectangles are taken from the entire figure ABKLIE, which is equivalent to AB’+BC”, there will evidently remain the square ACDE. Therefore, the square described, &c.

Scholium. This proposition is expressed algebraically thus :

(0-6)=a-2ab+6'. Cor.

(a+b)-(ab)=4ab.

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The rectangle contained by the sum and difference of two lines, is equivalent to the difference of the squares of those lines

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Let AB, BC be any two lines; the rectangle contained by the sum and difference of AB and BC, is equivalent to the difference of the squares on AB and BC; that is,

(AB+BC) X(AB-BC)=AB’— BC?. Upon AB describe the square ABKF, F

к and upon AC describe the square ACDE; produce AB so that BI shall be equal to E!

H L BC, and complete the rectangle AILE.

The base Al of the rectangle AILE is the sum of the two lines AB, BC, and its altitude AE is the difference of the same

C lines; therefore AILE is the rectangle contained by the sum and difference of the lines AB, BC. But this rectangle is composed of the two parts ABHE and BILH; and the part BILH is equal to the rectangle EDGF, for BH is equal to DE, and BI is equal to EF. Therefore AILE is equivalent to the figure ABHDGF. But ABHDGF is the excess of the square ABKF above the square DHKG, which is the square of BC; therefore,

(AB+BC) X(AB-BC)=AB-BC'.

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