1. The square on the sum of two segments is equal to the sum of the squares on the segments and twice the rectangle on the segments.

α

b

2

a2

ab

(a+b)2= a2 + b2+2ab.

2. The rectangle on the sum and difference of two segments is equal to the difference of the squares on these segments. (a+b)(a − b) = a2 — b2.

3. The sum of the squares on the sum and on the difference of two segments is equal to twice the sum of the squares on the segments.

(a+b)2+(a - b)2=2(a2+b2), a >b.

4. The difference of the squares on the sum and on the difference of two segments is equal to four times the rectangle on the segments.

(a+b)2- (a - b)2=4ab, ab.

EXERCISES.

1. To prove 4 of Art. 166°.

Let AH-a and

α

ab

b

b

E

F

(a-b)2

ab

L

нь в

ab

G b

HB-6 be the segments, so that AB is their sum. Through H draw HG || to BC, a side of the Make HG=a, square on AB. and complete the square FGLE, as in the figure, so that FG is Ka-b.

Then AC is (a+b)2 and EG is (a-b); and their difference is

the four rectangles AF, HK, CL, and DE; but these each have a and b as adjacent sides.

(a + b)2 − (a − b)2=4ab.

2. State and prove geometrically (a - b)2 = a2+b2 — 2ab.

3. State and prove geometrically

(a+b)(a+c)=a2+a(b+c)+bc.

4. State and prove geometrically by superposition of areas

(a+b)2+(a−b)2+2(a+b)(a−b)=(2a), where a and b

denote segments.

5. If a given segment be divided into any three parts the square on the segment is equal to the sum of the

squares on the parts together with twice the sum of the rectangles on the parts taken two and two.

6. Prove, by comparison of areas from the Fig. of Ex. 1, that (a+b)2=2b(a+b)+2b(a−b)+(a - b)2, and state the theorem in words.

SECTION IV.

AREAL RELATIONS.

167°. Def.—1. The segment which joins two given points is called the join of the points; and where no reference is made to length the join of two points may be taken to mean the line determined by the points.

2. The foot of the perpendicular from a given point to a given line is the orthogonal projection, or simply the projection, of the point upon the line.

3. Length being considered, the join of the projection of two points is the projection of the

join of the points.

Thus if L be a given line and P, Q, two given points, and PP', QQ′ perpendiculars upon L; PQ is the join of P and Q, P' and Q' are the projections of P and Q upon L, and the segment P'Q' is the projection of PQ upon L.

Ρ

Q L

168°. Theorem.-The sum of the projections of the sides of

any closed rectilinear figure, taken in cyclic order with

B

A' B'

C D' L

respect to any line, is zero.

ABCD is a closed rectilinear figure and L is any line. Then

Pr.AB+ Pr.BC+ Pr.CD+ Pr.DA=0

Proof. Draw the perpendiculars AA', BB', CC', DD', and the sum of the projections becomes A'B' + B'C'+C'D' + D'A'.

But D'A' is equal in length to the sum of the three others and is opposite in sense. .. the sum is zero.

It is readily seen that since we return in every case to the point from which we start the theorem is true whatever be the number or disposition of the sides.

This theorem is of great importance in many investigations.

Cor. Any side of a closed rectilinear figure is equal to the sum of the projections of the remaining sides, taken in cyclic order, upon the line of that side.

Def. In a right-angled triangle the side opposite the right angle is called the hypothenuse, as distinguished from the remaining two sides.

E

169°. Theorem.-In any right-angled triangle the square on one of the sides is equal to the rectangle

G

J

Then ...

Also,

and

B

K F

on the hypothenuse and the projection of that side on the hypothenuse.

ABC is right-angled at B, and BD is ▲ HAC. Then AB2=AC. AD.

Proof. Let AF be the on AC, and let EH be || to AB, and AGHB be a □, since LB is a

Proof-AF is the on AC and AH

Now

i.e.,

☐AH=ABLE=□ADKE,
AB2=AC. AD.

(140°)

q.e.d.

As this theorem is very important we give an alternative

proof of it.

is the on AB, and BD is

H

AC.

Cor. 1. Since AB2=AC. AD we have from symmetry

... adding,

or

BC2 AC. DC,
AB2+BC2=AC(AD+DC),

AB2+BC2=AC2.

:. The square on the hypothenuse of a right-angled triangle is equal to the sum of the squares on the remaining sides.

This theorem, which is one of the most important in the whole of Geometry, is said to have been discovered by Pythagoras about 540 B.C.

Cor. 2. Denote the sides by a and c and the hypothenuse by b, and let a1 and c1 denote the projections of the sides a and cupon the hypothenuse.

Then

and

..

a2=a1b, c2=c1b,
a2+c2= b2.

Cor. 3. Denote the altitude to the hypothenuse by p.
Then b=c1+a1, and ADB and CDB are right-angled at D,

(166°, 1)

i.e., the square on the altitude to the hypothenuse is equal to the rectangle on the projections of the sides on the hypothenuse.

Def. The side of the square equal in area to a given rectangle is called the mean proportional or the geometric mean between the sides of the rectangle.

Thus the altitude to the hypothenuse of a right-angled ▲ is a geometric mean between the segments into which the altitude divides the hypothenuse. (169°, Cor. 3)

And any side of the▲ is a geometric mean between the hypothenuse and its projection on the hypothenuse. (169)

170°. Theorem.-If the square on one side of a triangle is equal to the sum of the squares on the remaining sides, the triangle is right-angled at that vertex which is opposite the side having the greatest square. (Converse of 169°, Cor.) If AC2= AB2+BC2, the LB is a 7

B

Proof. Let ADC be a O on AC.
AC2=AB2+BC2,

AB is < AC.

.. a chord AD can be found equal to AB. Then the AADC is right-angled at D.

171°. Theorem 169° with its corollaries and theorem 170° are extensively employed in the practical applications of Geometry. If we take the three numbers 3, 4, and 5, we