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Cor. 1. A triangle is equal to half a parallelogram of the same base and altitude, because the altitude is the perpendicular distance between the parallels, which is every where equal, by the definition of parallels.

Cor. 2. If the base of a parallelogram be half that of a triangle, of the same altitude, or the base of the triangle be double that of the parallelogram, the two figures will be equal to each other.

THEOREM XXVII.

Rectangles that are contained by equal lines, are equal to each other.

LET BD, FH, be two rectangles, having the sides AB, BC, equal to the sides EF, FG, each to each; then will the rectangle BD be equal to the rectangle FH.

For, draw the two diagonals AC, EG, dividing the two parallelograms each into two equal parts. Then the two triangles ABC, EFG, are equal to each other (th. 1), because they have the two sides AB, BC, and the con

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tained angle B, equal to the two sides EF, FG, and the contained angle F. (hyp.) But these equal triangles are the halves of the respective rectangles and because the halves, or the triangles, are equal, the wholes, or the rectangles DB, HF, are also equal (ax. 6).

Cor. The squares on equal lines are also equal; for every square is a species of rectangle.

THEOREM XXVIII.

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

LET AC be a parallelogram, BD a diagonal, EIF parallel to AB or DC, and GIH parallel to AD or BC, making AI, IC, complements to the parallelograms EG, HF, which are about the diagonal DB: then will the complement AI be equal to the complement IC.

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For, since the diagonal DB bisects the three parallelograms AC, EG, HG (th. 22); therefore, the whole triangle DAB being equal to the whole triangle DCB, and the parts DEI, IHB, respectively equal to the parts DGI, IFB, the remaining parts AI, IC, must also be equal (ax. 3).

THEOREM XXIX.

A trapezoid, or trapezium having two sides parallel, is equal to half a parallelogram, whose base is the sum of those two sides, and its altitude the perpendicular distance between them.

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LET ABCD-be the trapezoid, having its two sides AB, DC, parallel; and in AB produced take BE, equal to DC, so that AE may be the sum of the two parallel sides; produce DC also, and let EF, GC, BH, be all three parallel to AD. Then is AF a parallelogram of the same altitude with the trapezoid ABCD, having its base AE equal to the sum of the parallel sides of the trapezoid; and it is to be proved that the trapezoid ABCD is equal to half the parallelogram AF.

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Now, since triangles, or parallelograms, of equal bases and altitude, are equal (cor. 2, th. 25), the parallelogram DG is equal to the parallelogram HE, and the triangle CGB is equal to the triangle CHB; consequently the line BC bisects, or equally divides, the parallelogram AF, and ABCD is the half of it.

THEOREM XXX.

The sum of all the rectangles contained under one whole line, and the several parts of another line, any way divided, is equal to the rectangle contained under the two whole lines.

LET AD be the one line, and AB the other, divided into the parts AE, EF, FB; then will the rectangle contained by AD and AB, be equal to the sum of the rectangles of AD and AE, and AD and EF, and AD and FB: thus expressed, AD.AB = AD.AE + AD.EF + AD.FB.

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For, make the rectangle AC of the two whole lines AD, AB; and draw EG, FH, perpendicular to AB, or parallel to AD, to which they are equal (th. 22). Then the whole rectangle AC is made up of all the other rectangles AG, EH, FC: but these rectangles are contained by AD and AE, EG and EF, FH and FB; which are equal to the rectangles of AD and AE, AD and EF, AD and FB, because AD is equal to each of the two EG, FH. Therefore the rectangle AD.AB is equal to the sum of all the other rectangles AD.AE, AD.EF, AD.FB.

Cor. If a right line be divided into any two parts, the square on the whole line is equal to both the rectangles of the whole line and each of the parts.

THEOREM XXXI.

The square of the sum of two lines is greater than the sum of their squares, by twice the rectangle of the said lines. Or, the square of a whole line is equal to the squares of its two parts, together with twice the rectangle of those parts.

LET the line AB be the sum of any two lines AC, CB; then will the square of AB be equal to the squares of AC, CB, together with twice the rectangle of AC.CB. That is, AB2: AC2+ CB2 + 2AC.CB.

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For, let ABDE be the square on the sum or whole line AB, and ACFG the square on the part AC. Produce CF and GF to the other sides at H and I.

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From the lines CH, GI, which are equal, being each equal to the sides of the square AB or BD (th. 22), take the parts CF, GF, which are also equal, being the sides of the square AF, and there remains FH equal to FI, which are also equal to DH, DI, being the opposite sides of the parallelogram. Hence the figure HI is equilateral: and it has all its angles right ones (cor. 1, th. 22); it is therefore a square on the line FI, or the square of its equal CB. Also the figures EF, FB, are equal to two rectangles under AC and CB, because GF is equal to AC, and FH or FI equal to CB: but the whole square AD is made up of the four figures, viz. the two squares AF, FD, and the two equal rectangles EF, FB; that is, the square of AB is equal to the squares of AC, CB, together with twice the rectangle of AC, CB.

Cor. Hence, if a line be divided into two equal parts; the square of the whole line will be equal to four times the square of half the line.

THEOREM XXXII.

The square of the difference of two lines is less than the sum of their squares, by twice the rectangle of the said lines.

LET AC, BC, be any two lines, and AB their difference : then will the square of AB be less than the squares of AC, BC, by twice the rectangle of AC and BC. Or AB2 = AC2 + BC2 — 2AC.BC.

For, let ABDE be the square on the difference AB, and ACFG the square on the line AC. Produce ED to H; also produce DB and HC, and draw KI, making BI the square of the other line BC.

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Now, it is obvious, that the square AD is less than the two squares AF, BI, by the two rectangles EF, DI: but GF is equal to the one line AC, and GE or FH is equal to the other line BC; consequently the rectangle EF, contained under EG and GF, is equal to the rectangle of AC and BC.

Again, FH being equal to CI or BC or DH, by adding the common part HC, the whole HI will be equal to the whole FC, or equal to AC; and consequently the figure DI is equal to the rectangle contained by AC and BC.

Hence the two figures EF, DI, are two rectangles of the two lines AC, BC; and consequently the square of AB is less than the squares of AC, BC, by twice the rectangle AC, BC.

THEOREM XXXIII.

The rectangle under the sum and difference of two lines, is equal to the difference of the squares of those lines.*

LET AB, AC, be any two unequal lines; then will the difference of the squares of AB, AC, be equal to a rectangle under their sum and difference: that is, AB2 (AB + AC) (AB — AC).

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For, let ABDE be the square of AB, and ACFG the square of AC. Produce DB till BH be equal to AC; draw HI parallel to AB or ED, and produce FC both ways to I and K.

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Then the difference of the two squares AD, AF, is evidently the two rectangles EF, KB. But the rectangles EF, BI, are equal, being

* This and the two preceding theorems are evinced algebraically, by the three expressions (a + b)2 = a2 + 2ab+b2 = a2 + b2+2ab

(a - b)2 = a2 — 2ab+b2 = a2 + b2 — 2ab

(a + b) (a - b) = a2 — b2,

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Of course it is here assumed that an algebraic product corresponds to a geometrical rectangle, which is shown in the Application of Algebra to Geometry.

contained under equal lines; for EK and BH are each equal to AC, and GE is equal to CB, being equal to the difference between AB and AC, or their equals AE and AG. Therefore the two EF, KB, are equal to the two KB, BI, or to the whole KH; and consequently KH is equal to the difference of the squares AD, AF. But KH is a rectangle contained by DH, or the sum of AB and AC, and by KD, or the difference of AB, and AC: therefore the difference of the squares of AB, AC, is equal to the rectangle under their sum and difference.

THEOREM XXXIV.

In any right-angled triangle, the square of the hypothenuse is equal to the sum of the squares of the other two sides.

LET ABC be a right-angled triangle, having the right angle C; then will the square of the hypothenuse AB, be equal to the sum of the squares of the other two sides AC, CB. Or AB AC + BC.

For, on AB describe the square AE, and on AC, CB, the squares AG, BH; then draw CK parallel to AD or BE; and join AI, BF, CD, CE.

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Now, because the line AC meets the two CG, CB, so as to make two right angles, these two form one straight line GB (cor. 1, th. 6). And because the angle FAC is equal to the angle DAB, being each a right angle, or the angle of a square; to each of these angles add the common angle EAC, so will the whole angle or sum FAB, be equal to the whole angle or sum CAD: but the line FA is equal to the line AC, and the line AB to the line AD, being sides of the same square; so that the two sides FA, AB, and their included angle FAB, are equal to the two sides CA, AD, and the contained angle CAD, each to each: therefore the whole triangle AFB is equal to the whole triangle ACD (th. 1).

But the square AG is double the triangle AFB, on the same base FA, and between the same parallels FA, GB (th. 26); in like manner the parallelogram AK is double the triangle ACD, on the same base AD, and between the same parallels AD, CK: and since the doubles of equal things are equal (ax. 6); therefore the square AG is equal to the parallelogram AK.

In like manner, the other square BH is proved equal to the other parallelogram BK: consequently the two squares AG and BH together, are equal to the two parallelograms AK and BK together, or to the whole square AE; that is, the sum of the two squares on the two less sides is equal to the squares on the greatest side.

Cor. 1. Hence, the square of either of the two less sides, is equal to the difference of the squares of the hypothenuse and the other side (ax. 3); or equal to the rectangle contained by the sum and difference of the said hypothenuse and other side (th. 33).

Cor. 2. Hence, also, if two right-angled triangles have two sides of the one equal to two corresponding sides of the other; their third sides will also be equal, and the triangles identical.

THEOREM XXXV.

In any triangle, the difference of the squares of the two sides is equal to the difference of the squares of the segments of the base, or of the two lines, or distances, included between the extremes of the base and the perpendicular.

LET ABC be any triangle, having CD perpendicu lar to AB; then will the difference of the squares of AC, BC, be equal to the difference of the squares of AD, BD; that is, AC2 - BC2 = AD2 . BD2.

For, since (th. 34) AC2 = = AD2+ DC2 and BC2 = BD2 + DC2, the differences of these are equal; that is, AC2 BC2 = AD2

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BD2.

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Cor. The rectangle of the sum and difference of the two sides of any triangle, is equal to the rectangle of the sum and difference of the distances between the perpendicular and the two extremes of the base, or equal to the rectangle of the base and the difference or sum of the segments, according as the perpendicular falls within or without the triangle.

That is, (AC+ BC) (AC — BC) = (AD + BD) (AD or, (AC + BC) (AC — BC) = AB (AD

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BD) in the 2nd fig. ; and (AC + BC) (AC — BC) = AB (AD + BD) in the 1st fig.

THEOREM XXXVI.

In any obtuse-angled triangle, the square of the side subtending the obtuse angle, is greater than the sum of the squares of the other two sides, by twice the rectangle of the base and the distance of the perpendicular from the obtuse angle.

LET ABC be a triangle, obtuse angled at B, and CD perpendicular to AB; then will the square of AC be greater than the squares of AB, BC, by twice the rectangle of AB, BD: that is, AC2 = AB2+ BC2 + 2AB. BD. See the 1st fig. above, or below.

For, AD2:

= AB2+ BD2 + 2AB. BD (th. 31),

and AD2 + CD2 = AB2 + BD2 + CD2 + 2AB . BD (ax. 2):
But AD2 + CD2 = AC2, and BD2 + CD2 = BC2 (th. 34);
therefore AC2 = AB2 + BC2 + 2AB. BD.

THEOREM XXXVII.

In any triangle, the square of the side subtending an acute angle, is less than the squares of the base and the other side, by twice the rectangle of the base and the distance of the perpendicular from the acute angle.

LET ABC be a triangle, having the angle A acute, and CD perpendicular to AB; then will the square of BC be less than the squares of AB, AC, by twice the rectangle of AB, AD. That is, BC2 + 2AD. DB = AB2 + AC2.

For BD2 = AD2 + AB2 - 2AD. AB (th. 32),

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and BD2 + DC2 = AD2 + DC2 + AB2 - 2AD. AB (ax. 2);
therefore BC2 AC2 + AB2 · 2AD. AB (th. 34).

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