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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. Zs 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 2s of (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. 2 has all its 2s rt. Zs.

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

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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.); . . iu the AS AEH, BEF, the two EA, AH = FB, BE, each to each, and the rt. / EAH = rt. | EBF; ... the base EH =

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, 4 HEF = | EAH (Ax. 3) = a rt. L. (Hyp.)

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

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.

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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 squares

described ироп the sides which contain the right angle.

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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 (= ABP + AC2).

Const. On describe the squares BDEC, and on AB, AC describe the squares GB, HC (Prop.XÀVI.); 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 Z $ = 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. Z DBC = rt. Z FBA (Ax. 11), add to each the 2 ABC; .. the 2 DBA = Z FBC.

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

= base FC, and the A ABD = A FBC. (Prop. IV.)

5. But the om bl = 2 A ABD, upon the same base bd, and between same || S AL, BD; and the square GB = 2 - 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,

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be

proved, by joining AE, KB, that the omcL = square sc.

8. Hence, by addition, the square upon BC = sum of the squares upon AB and ac, or bc2 = AB2 + AC2.

Wherefore, in any rt. 2d A, 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. 2d A, is called the hypothenuse, from the verb únoteivelv, to subtend.

PROP. CCC. THEOR.

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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 a rectangular am, whose diam. is BD; then BD2 DC? + Bc.

DEMONST.–For ABD is a rt. 4d A; .. BD2

AB? + AD? = DC2 + B! Bc2. (Prop. XLVII.)

Wherefore the square &c.

Cor.-If the Om be a square, the square upon the diameter is equal to twice the square upon either of the sides ; for in this case AB = AD; ,. BD2 = ABP + ADP 2 AB2.

PROP. DDD. THEOR.

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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. 20 A, having the rt. LABC. 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? AB? + Bc2 (Prop. XLVII.); add bdo to each side; ::. Ac? + BD?

AB? + BD? + BC (Ax. 2.) AD2 + BC%. (Prop. XLVII.)

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

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PROP. EEE. THEOR.

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 BD and dc.

DEMONST.-Because ADC is a rt.
ZA, .. ACP AD2 + Dc; and
because ABD is a rt. 41 A, .. AB?
AD2 + BD?; . ., by subtraction, B

AB? DC2 BD2.
Wherefore in any A, if a line &c.-Q. E. D.

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