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.

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A

Let A B C be a right triangle with its right angle at C.

We are to prove À Co + C B = A B2
Draw Col to A B.
Then
Ā = AOX AB,

$ 289 (the square on a side of a rt. A is equal to the product of the hypotenuse by the adjacent segment made by the I let fall from the vertex of the rt. 2); and B (12 = BOX AB,

$ 289 By adding, À CP + BT= (A 0 + BO) A B,

= A B X A B,
= A B?.

Q. E. D.

332. COROLLARY. The side and diagonala
of a square are incommensurable.
Let A B C D be a square, and A C the

diagonal.
Then A B + BC = AC?.
or,

2 A Bo = A C".
Divide both sides of the equation by A B?,

B

Extract the square root of both sides the equation,
then

А С
ĀB =V2.

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 inconimensurable 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.

LE
Let A B C be a right A, having the right angle BAC.

We are to prove BC“ = B A' + AC?.
On B C, C A, A B construct the squares B E, CH, A F.

Through A draw A L II to C E.

Draw A D and FC.
ZB A C is a rt. 2,

Hyp.
ZBA G is a rt. 2,

Cons.
.:CA G is a straight line.
Also
Z C A H is a rt. 2,

Cons.
.: BA H is a straight line.
Now
2 DBC= Z FBA,

Cons, (each being a rt. 2).

and

Add to each the ABC;
then

LABD=ZFBC,
A ABD=AFBC.

§ 106
Now DBL is double A ABD,
(being on the same buse B D, and between the same lls, A L and BD),

and square A F is double A FBC
(being on the same base FB, and between the same lls, FB and G C);

OBL= square A F. In like manner, by joining A E and B K, it may be proved that

OCL= square CH.
Now the square on BC= O BL+O CL,

= square A F + square C H,
.. BOʻ = BĀ+ A TP.

Q. E. D.

ON PROJECTION. 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.

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RB A

DB Fig. 1.

Fig. 2. The Projection of a Finite Straight Line, as C D (Fig. 1), upon a straight line of indefinite length, as A B, is the part of the line A B intercepted between the perpendiculars C P and DR, let fall from the extremities of the line C D.

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

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

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.

Fig. 1.

Fig. 2. Let C be an acute angle of the triangle A BC, and

DC the projection of A Cupon BC.

We are to prove A B' = B C + ACC – 2 BC XD C.
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 Bʻ = B C + D C – 2 BC X DC.

Add A D' to both sides of the equality;
then, A D’ + D Bo = BC + A + D – 2 BC X DC.
But
A D + D B = AB,

$ 331 (the sum of the squares on two sides of a rt. A is equivalent to the square

on the hypotenuse) ;
and
AD + D C = A C",

$ 331 Substitute A B’ and Ā for their equivalents in the above equality; then, A B = B 0° + A C – 2 BCX DC.

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.

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Let C be the obtuse angle of the triangle A B C, and

CD be the projection of A C upon B C produced.
We are to prove AB* = B C + A + 2 BCX DC.

DB= BC + DC.
Squaring, DB° = B C + D (4 + 2 B C X DC.

Add A D to both sides of the equality; then, AD + DB° = B C + A D’+ D C + 2 B C X DC.

AD? + D B = A B”, (the sum of the squares on two sides of a rt. A is equivalent to the square

on the hypotenuse);
AD? + D C = A C?..

§ 331 Şubstitute Ā B and Ā C for their equivalents in the above equality; then, A Bʻ = B C + A C + 2 BCX DC.

Q. E. D.

But

and

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.