Page images
PDF
EPUB

PROPOSITION VIII. THEOREM.

331. The sum of the squares described on the two sides of a right triangle is equivalent to the square described on the hypotenuse.

C

[blocks in formation]

Let ABC be a right triangle with its right angle at C.

We are to prove AC2 + CB2= A B2

Ᏼ :

=

ACAOX AB,

Draw COL to A B.

Then

$289 (the square on a side of a rt. ▲ is equal to the product of the hypotenuse by the adjacent segment made by the 1 let fall from the vertex of the rt. 4);

and

BC12

=

ВОХАВ,

By adding, AC+ BC2= (AO+BO) A B,

$ 289

[blocks in formation]

332. COROLLARY. The side and diagonal Α

of a square are incommensurable.

Let ABCD be a square, and AC the

[blocks in formation]

Divide both sides of the equation by A B2,

A C2

= 2.

A B

[blocks in formation]

Extract the square root of both sides the equation,

[blocks in formation]

Since the square root of 2 is a number which cannot be exactly found, it follows that the diagonal and side of a square are two incommensurable lines.

ANOTHER DEMONSTRATION.

333. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides.

G

[blocks in formation]

Let ABC be a right ▲, having the right angle BAC.

[blocks in formation]

On BC, CA, A B construct the squares B E, CH, A F.

Through A draw A L to CE.

Draw A D and FC.

[merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]
[merged small][merged small][merged small][ocr errors][merged small][merged small][merged small]

(being on the same base BD, and between the same ||s, AL and B D), and square A F is double ▲ FBC,

(being on the same base FB, and between the same |\s, FB and G C); .. □ B L = square A F.

In like manner, by joining A E and B K, it may be proved

[merged small][ocr errors][merged small][merged small][merged small][merged small][merged small][merged small][merged small]

334. DEF. The Projection of a Point upon a straight line of indefinite length is the foot of the perpendicular let fall from the point upon the line. Thus, the projection of the point C upon the line A B is the point P.

[blocks in formation]

The Projection of a Finite Straight Line, as CD (Fig. 1), upon a straight line of indefinite length, as A B, is the part of the line AB intercepted between the perpendiculars CP and D R, let fall from the extremities of the line C D.

Thus the projection of the line CD upon the line A B is the line P R.

If one extremity of the line CD (Fig. 2) be in the line A B, the projection of the line CD upon the line A B is the part of the line AB between the point D and the foot of the perpendicular CP; that is, D P.

PROPOSITION IX. THEOREM.

335. In any triangle, the square on the side opposite an acute angle is equivalent to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side.

[blocks in formation]

Let C be an acute angle of the triangle ABC, and DC the projection of AC upon BC.

We are to prove

A B2 = B C2 + A C2 − 2 B C × DC.

If D fall upon the base (Fig. 1),

DB BC-DC;

If D fall upon the base produced (Fig. 2),

DB DC - BC.

In either case D B2 = B C2 + D C2 – 2 B C × D C.

[blocks in formation]
[ocr errors]

then, A D2 + DB2 = B C2 + A D2 + D C2 − 2 B C × DC.

But

AD2 + D B2 = A B2,

$331

(the sum of the squares on two sides of a rt. ▲ is equivalent to the square

and

on the hypotenuse);

AD2 + DO2 = A C2,

§ 331

Substitute A B2 and AC2 for their equivalents in the above equality;

then, AB2 = B C2 + A C2 − 2 B C × D C.

Q. E. D.

PROPOSITION X. THEOREM.

336. In any obtuse triangle, the square on the side opposite the obtuse angle is equivalent to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other on that side.

[blocks in formation]

Let C be the obtuse angle of the triangle ABC, and CD be the projection of AC upon BC produced.

We are to prove

A B2 = B C2 + A T2 + 2 B C X DC.

DB BC + D C.

Squaring, DB = BC2 + DC2 + 2 B C X DC.

×

Add AD to both sides of the equality;

then, AD2 + DB = BC2+ AD2 + DC2 + 2 BCX DC.

But

A D2 + D B2 = A B2,

$331

(the sum of the squares on two sides of a rt. ▲ is equivalent to the square

and

on the hypotenuse);

A D2 + D C2 = A C2.

§ 331

Substitute AB and AC for their equivalents in the above equality;

then,

Ꭺ Ᏼ = B C2 + A C2 + 2 B C × DC.

Q. E. D.

337. DEFINITION. A Medial line of a triangle is a straight line drawn from any vertex of the triangle to the middle point of the opposite side.

« PreviousContinue »