(5. 4;) hence the area of the triangle must be BC ×AD, or BCXAD. Therefore, The area, &c.

Cor. Two triangles of the same altitude, are to each other as their bases; and two triangles of the same base, are to each other as their altitudes.

PROPOSITION VII. THEOREM.

The area of the trapezium ABCD, is equal to its altitude CE, multiplied by half the sum of its parallel bases, AB, CD; that is, ABCD=CEX (AB+CD).

Join AC. Now since CE > AB is the measure of the triangle ABC; (6.4;) and since CEx DC, is the measure of the triangle ACD, hence

D

the measure of both triangles is CE× (AB+CD). But the sum

A

E B

of the two triangles ABC+ACD, is equal to the trapezium ABCD; (Ax. 13;) therefore the area of ABCD= CEX (AB+CD). (Ax. 1.) Hence,

The area of every trapezium is equal to the product of its altitude into half the sum of its parallel bases.

If a line AC is divided into two parts AB, BC, the square described on the whole line AC, is equivalent to the sum of the squares described on each of the parts AB, BC, added to twice the rectangle contained by those parts. That is,

AC3, or (AB+BC)2=AB2+BC2+2 (AB+BC).

Construct the square ACDE; take AF=AB; draw FG parallel to AC, and BH parallel to AE.

The square ACDE is divided into

four parts.
The first, ABIF is the
square described on AB, and is equal to F
AB2; (5. 4. Cor. 1 ;) for we made AF
=AB. The second, GDHI is the
square described on BC; for since AC

E

A

B

C

= AE and AB AF, hence AC-AB= AE-AF, which gives BC=EF; and since FG is parallel to ED, and BH to CD, therefore DG EF and IG=BC; hence GDHI=BC2.

The third and fourth parts of the square are the rectangles, BG and FH. And since IB AB each being the side of a square, the measure of the rectangle BG, is ABX BC; and since FI AB, and IH=IG=BC, the measure of the rectangle FH, is AB × BC; hence the sum of the two rectangles BG+FH=2(ABXBC). Therefore, (Ax. 10,)

AC AB+BC2+2 (AB × BC). Hence,

If a line is divided into any two parts, the square described on the whole line is equivalent to the sum of the squares described on each of the parts, added to twice the rectangle contained by those parts.

Scholium 1. This property is the same as that demonstrated in algebra for obtaining the square of a binomial, (Arts. 173 and 398,) which is thus expressed:

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

Scholium 2. If a straight line AC is bisected, the square described on the whole line is equivalent to four times the square described on half the line. This explains the apparent paradox in arithmetic, that the square of a half is equal to a fourth. For the

square

792538 A

of AB, the half of AC, is equivalent to a fourth of the square of the whole AC.

If AB, BC, be two lines and AC their difference, the square described on AC, will be equivalent to the square of AB, plus the square of BC, minus twice the rectangle contained by AB and BC; that is,

AC2, or (AB-BC)'=AB+BC2-2 (ABX BC).

figure ABILKE=(AC2+CBIG+GLKD)=(AB2+BC2). But the two rectangles CI, GK, are each measured by ABX BC, and the sum of both of the rectangles is measured by 2 (ABX BC); taking these two rectangles from each member of the equation we have AC2=AB2+BC2—2(AB× BC). Hence,

The square described on the difference of two lines, is equivalent to the sum of the squares described on each of the lines, minus twice the rectangle contained by those lines.

Scholium. This proposition is equivalent to the algebraical formula, (a—b)2=a2-2ab+b2. (Art. 173.)

The rectangle contained by the sum and the difference of two lines, AB, BC, is equivalent to the difference of the squares of those lines; that is,

(AB+BC)×(AB-BC)=AB3-BC2.

On AB and AC, construct the squares ABIF, ACDE; produce AB till the produced part BK, is E equal to BC; and complete the rectangle AKLE.

The base AK of the rectangle

F

G I

H

D

AL is the sum of the two lines A

св K

AB, BC; its altitude AE is the difference of the same lines; therefore the rectangle AKLE is equal to (AB+BC)× (AB-BC). But this rectangle is composed of the two parts ABHE+BHLK; and the part BHLK is equal to the rectangle EDGF, because BH is equal to DE, and BK to EF; hence AKLE is equal to ABHE+EDGF. These two parts make up the square ABIF minus the square DHIG, which latter is the square described on BC: therefore

(AB+BC) × (AB-BC)=AB2-BC. Hence,

The rectangle contained by the sum and difference of two lines, is equivalent to the difference of the squares of those lines. Scholium. This proposition is equivalent to the algebraical formula, (a+b)× (a—b)=a2-b2. (Art. 191. )

PROPOSITION XI. THEOREM.

The square described on BC the hypothenuse of a right-angled triangle ABC, is equivalent to the sum of the squares described on the two sides, AB, AC.

made up of the same angle ABC, together with the rightangle ABH; hence the angle ABF is equal to HBC. But we have AB BH, being sides of the same square; and BF= BC, for the same reason: therefore the triangles ABF, HBC have two sides and the included angle in each equal; therefore they are themselves equal. (5. 1.)

The triangle ABF is half of the rectangle BE, because they have the same base BF, and the same altitude BD. (2. 4.) The triangle HBC is in like manner half of the square AH: for the angles BAC, BAL being both right-angles, AC and AL form one and the same straight line parallel to HB; (3. 1;) and consequently the triangle HBC, and the square AH, which have the common base BH, have also the common altitude AB; hence the triangle is half of the square.

But the triangle ABF has already been proved equal to the triangle HBC; hence the rectangle BDEF, which is double of the triangle ABF, must be equivalent to the square AH, which is double of the triangle HBC. In the same manner it may be proved, that the rectangle CDEG is equivalent to the square AI. But the two rectangles BDFE, CDEG, taken together, make up the square BCGF: therefore the