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with the square on the other part, is equivalent to the square on the line which is made up of the whole sect and the first part.

301. By the associative and commutative laws, and 294,

([a + 6] + a)2 + b2 = (b + 2a)2 + 62

= 62 + 4ab + 4a + b2 = 2a2 + 4ab + 262 + 2a2.

By the distributive law, and 294,

2a2 + 4ab + 2b2 + 2a = 2(a2 + 2ab +62) + 2a2 = 2(a + b)2 + 2a?,

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Therefore

If a sect be divided into two equal and also two unequal parts, the squares on the two unequal parts are together double the squares on half the sect and on the line between the points of section.

302. By 294, and the associative and distributive laws,

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Therefore

If a sect be bisected and produced to any point, the square on the whole line thus produced and the square on the part of it produced are together double the squares on half the sect and on the line made up of the half and the part produced.

303. The projection of a point on a line is the foot of the perpendicular from the point to the line.

304. The projection of a sect upon a line is the part between the perpendiculars dropped upon the line from the ends of the sect.

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For example, A'B' is the projection of the sect AB on the

line c.

THEOREM I.

305. In an obtuse-angled triangle, the square on the side opposite the obtuse angle is greater than the sum of the squares on the other two sides by twice the rectangle contained by either of those sides and the projection of the other side upon it.

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HYPOTHESIS. A ABC, with X CAB obtuse.
CONCLUSION. a = b2 + 2 + 2bj.
PROOF. By 294,

(6 + 7) = 12 + 2bj + js. Adding h2 to both sides,

(b + ;)2 + ha = b2 + 2bj + j? + h2. But

(6 + ;)? + h = a?, and ja + h2 = ca, (243. In a right triangle, the square of the hypothenuse equals the sum of the

squares of the other sides.) ... a = 12 + 2bj + c2.

THEOREM II.

306. In any triangle, the square on a side opposite any acute angle is less than the sum of the squares on the other two sides by twice the rectangle contained by either of those sides and the projection of the other side upon it.

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HYPOTHESIS. A ABC, with X C acute.
CONCLUSION. + 2bj = ax + b2.
PROOF. By 295,

B2 + y2 = 2bj + AD.
Adding ha to both sides,

b2 + y2 + h = 2bj + AD + ha. But j? + h = a, and

AD + h2 = 6, (243. In a right triangle, the square of the hypothenuse equals the sum of the

squares of the other sides.)

.: 62 + a2 = 2bj + c.

..

...

307. Having now proved that in a triangle, By 243, if X. A = rt. x,

al = 12 + 2; By 305, if XA >rt. ,

al > 82 + 6; By 306, if * A < rt. *,

az < 82 + 62. Therefore, by 33, Rule of Inversion, If a? = 52 + 6,

= rt. * ; If aż > 82 + 6, .. XA > rt. * ; If a< 32 + 6, .. * A < rt. 4.

*

THEOREM III.

308. In any triangle, if a medial be drawn from the vertex to base, the sum of the squares on the two sides is equivalent to twice the square on half the base, together with twice the square on the medial.

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HYPOTHESIS. A ABC, with medial BM = i.
CONCLUSION. a2 + C2 =

2(16)2 + 222.
PROOF. CASE I. If X BMA = * BMC, then

a = (18)” + is, and (= (16)2 + 12. (243. In a right triangle, the square of the hypothenuse equals the sum of the

squares of the other sides.)

az =

CASE II. If X BMA does not equal XBMC, one of them must be the greater.

Call the greater BMC.
Then, in the obtuse-angled triangle BMC, by 305,

(16)2 + 12 + 2(36j). In A BMA, by 306,

C + 2({bj) = (16)2 + i2. Adding,

az + 6 + 2(16) = 2(18)2 + 2i2 + 2({bi),

:: a2 + c = 2(18)2 + 2i2.

309. COROLLARY. The difference of the squares on the two sides is equivalent to twice the rectangle of the base and the projection of the medial on it.

THEOREM IV.

310. The sum of the squares on the four sides of any quadrilateral is greater than the sum of the squares on the diagonals by four times the square on the sect joining the mid-points of the diagonals.

B

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HYPOTHESIS. EF is the sect joining the mid-points of the diagonals AC and BD of the quadrilateral ABCD. CONCLUSION. AB + BC2 + CD +

= AC + BD + 4EF2. PROOF. Draw BE and DE. By 308, AB + BC

+ 2BE, and

A
CD + DA

+ 2DE.
Adding,
AB + BC + CD + =

AC2

+

2

(499

BE + DE = -(BP)* + 2EF", ... AB + BC* + CD + DA = 4(45)* ++(3D + 4ET

But, by 308,

BD12

ACP + BD + 4EF.

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