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PROP. XLVI. PROB.

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GEN. ENUN.- To describe a square upon a given straight line.

PART. ENUN. — Let AB be the gn. st. line; then it is required to describe a square upon

it. Const. and DEMONST. 1. From the pt. A draw ac at rt. Z to AB (Prop. XI.); make Ad= AB (Prop. III.); through a draw DE || to AB, and through B draw BE || to AD (Prop. XXXI.) ; .:. ABED is a om; and ... AB = DE, and AD = BE. (Prop. XXXIV.)

2. But AB = AD (Const.) ; .-. AB = AD = EB = ED (Ax. 1). Hence ABED is equilateral.

3. And because AD meets the || st. lines AB, DE, ... the ZS BAD + ADE = 2 rt. Zs (Prop. XXIX.); but Z BAD is a rt. 2 (Const.); ... also 2 ADE is a rt. Z.

4. Now the opposite 28 of oms are = (Prop. XXXIV.); ... each of the opposite 4S ABE, BED is a rt. Z.

5. Hence the figure ABED is equiangular as well as equilateral; it is .. a square (Def. 30), and it has been described

upon

the
gn.

line AB.-Q. E. F.

Cor.—Hence every om that has one rt. _ has all its Zs rt. Zs.

Although not immediately deducible from the above Proposition, the following Theorem may be inserted here.

PROP. BBB. THEOR.

А

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Gen. Enun.-If in each of the sides of a square, a point be taken at equal distances from the four angles, the straight lines which join them will also form a square.

PART. ENUN.- Let ABCD be a square. E, F, G, H, · points at = distances from the four 28. Join EF, FG, GH, HE. Then the figure Ergh is also a square.

DEMONST.-1. Since AE = BF, and Ah=EB (Const.); .. in the AS AEH, BEF, the two EA, AH = FB, BE, each to each, and the rt. Z EAH

rt. | EBF; ... the base EH = base EF, and the L BEF = LAHE. (Prop. IV.)

2. In like manner, Fg and GH = EF or EH.

3. Hence EF = FG = GH = HE (Ax. 1), and ... the figure EFGA is equilateral.

4. Again, the exterior / HEB of A AEH = interior Ls EAH + AHE (Prop. XXXII.), and the / BEF = L AHE; .., by subtraction, / HEF = LEAH (Ax. 3) = a rt. 4. (Hyp.)

5. In the same manner the 28 at F, G, H may be shewn to be rt. $.

6. Hence the figure EFGH is also equiangular, and ... it is a square (Def. 29).

Wherefore, if in each of the sides &c.-Q. E. D.

B

E

E

PROP. XLVII. THEOR.

GEN. ENUN.-In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the

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the sides which contain the right angle.

squares described

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H

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BC

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E

Part. Enun.—Let ABC be a rt. Zda, having the rt. Z BAC; then the square described upon BC the sum of the squares upon AB and Ac (= AB2 + AC2).

CONST. — On describe the squares BDEC, and on AB, AC describe the squares GB, HC (Prop.XLVI.); through a draw AL || to BD or CE (Prop. XXXI.), and join Ad, FC (Post. 1).

DEMONST.-1. Now, because each of the <$ BAC, BAG is a rt. 2 (Const.) . . at the pt. A in the st. line AB, the line ca, Ag, on opposite side of it, makes the adjacent Zs to two rt. <s; and .:. ca is in the same st. line with AG. (Prop. XIV.)

2. For the same reason AB and Ah are in the same st. line.

3. Again, because the rt. < DBC = rt. Z FBA (Ax. 11), add to each the < ABC; ..

... the Z DBA = Z FBC.

4. Hence, in the AS ABD, FBC, the two AB, BD = FB, BC, each to each (Const.), and the included 2 ABD = included L FBC; ... the base AD

base Fc, and the A ABD = A FBC. (Prop. IV.) 5. But the om BL = 2 A ABD, upon the

BD,

and between same 11 % AL, BD;

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same base

and the square GB

= 2 A FBC, upon the same base FB,

and between same || SPB, GC. (Prop. XLI.)

6. Now the doubles of = things are = to one another (Ax. 6); .-. the om BL = square

GB.

7. In the same manner, it may be proved, by joining AE, KB, that the o mcL= square hc.

8. Hence, by addition, the square upon BC = sum of the squares upon AB and ac, or BC2 = AB2 + AC2. Wherefore, in any rt. Zd the

square

&c. -Q.E.D.

This Proposition is no less remarkable for its elegance than for its extensive utility. A few deductions are added for the student's consideration, who will observe, that the side which subtends the rt. 7 in a rt. 20 A, is called the hypothenuse, from the verb 'noteively, to subtend.

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PROP. CCC. THEOR.

A

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Gen. Enun.- The square described upon the diameter of a rectangular parallelogram, is equal to the square described upon two of its unequal sides.

Part. ENUN.-Let ABCD be a rectangular om, whose diam. is BD; then BD2 Dc2 + Bc.

DEMONST.–For ABD is a rt. 7 d
A; .. BD2 AB? + AD? Dc2 + B
Bc2. (Prop. XLVII.)

Wherefore the square &c.

CoR.-If the am be a square, the square upon the diameter is equal to twice the square upon either of the

for in this case AB = AD; ... BD2 = ABP + AD2 2 AB2.

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sides ;

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PROP. DDD. THEOR.

A

Gen. Enun.-If from one of the acute angles of a rightangled triangle, a line be drawn to the opposite side, the squares of that side and the line so drawn are, together, equal to the squares of the segment adjacent to the right angle and the hypothenuse.

Part. Enun.--Let ABC be a rt. 4d A, having the rt. | ABC. From A draw AD to any pt. in the opposite side bc; then the squares upon AD and BC = the squares upon ac and BD.

DEMONST.For since Ac + Bc2 (Prop. XLVII.); add Bd to each side; ::. ACC + BD2

AB? + BD? + BC (Ax. 2.) = ADP + Bc. (Prop. XLVII.)

Wherefore, if from one &c.Q. E. D.

AB?

B

PROP. EEE. THEOR.

A

Gen. Enun.-In any triangle, if a line be drawn from the vertex perpendicular to the base, the difference of the squares of the sides is equal to the difference of the squares of the segments of the base.

Part. ENUN.-Let ABC be any A; from the vertex A draw AD I to the base BC; then the difference of the squares upon AB and AC

difference of squares upon bp and dc.

DEMONST.-Because ADC is a rt. 24,.. AC AD? + Dc?; and because ABD is a rt. 2d A, .. ABC

AD2 + BD2; ..., by subtraction, B AC? – AB? = DC? - BD2.

Wherefore in any A, if a line &c.-Q. E. D.

L

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