PROPOSITION VIII. THEOREM.

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

Let AC be the line, and B the point of division; then, is
AC or (AB+BC)2=AB2 + BC2+2AB × BC.
Construct the square ACDE; take AF E
AB; draw FG parallel to AC, and BH par-
allel to AE.

F

H

D

B C

The square ACDE is made up of four parts; the first ABIF is the square described on AB, since we made AF-AB: the second IDGH is A the square described on IG, or BC; for since we have AC= AE and AB=AF, the difference, AC-AB must be equal to the difference AE-AF, which gives BC=EF; but IG is equal to BC, and DG to EF, since the lines are parallel; therefore IGDH is equal to a square described on BC. And those two squares being taken away from the whole square, there remains the two rectangles BCGI, EFIH, each of which is measured by ABX BC: hence the large square is equivalent to the two small squares, together with the two rectangles.

Cor. If the line AC were divided into two equal parts, the two rectangles EI, IC, would become squares, and the square described on the whole line would be equivalent to four times the square described on half the line.

Scholium. This property is equivalent to the property demonstrated in algebra, in obtaining the square of a binominal; which is expressed thus:

(a+b)2=a2+2ab - +12.

PROPOSITION IX. THEORFM.

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

Let AB and BC be two lines, AC their difference; then is AC, or (AB-BC)2=AB+BC2-2AB x BC.

G I

Describe the square ABIF; take AE L F =AC; draw CG parallel to to BI, HK parallel to AB, and complete the square K EFLK.

E

PROPOSITION X. THEOREM.

The two rectangles CBIG, GLKD, are each measured by AB × BC; take them away from the whole figure ABILKEA, which is equivalent to AB2+BC, and there will evidently remain the square ACDE; hence the theorem is true.

A

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

D

The rectangle contained by the sum and the difference of two lines, is equivalent to the difference of the squares of those lines.

Let AB, BC, be two lines; then, will

(AB+BC) × (AB-BC)=AB2-BC2. On AB and AC, describe the squares ABIF, ACDE; produce AB till the produced part BK is equal to BC; and complete the rectangle AKLE.

E

C B K

The base AK of the rectangle EK, is the sum of the two lines 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 ARIF minus the square DHIG, which latter is equal to a square described on BC: hence we have

G*

G I

D

H

L

(AB+BC) × (AB—BC) = AB2—BC2.

Scholium. This proposition is equivalent to the algebraical formula, (a+b) x (a—b) = a2—b2.

PROPOSITION XI. THEOREM

The square described on the hypothenuse of a right angled tr angle is equivalent to the sum of the squares described on the other two sides.

Let the triangle ABC be right angled at A. Having described squares on the three sides, let fall from A, on the hypothenuse, the perpendicular AD, which produce to E; and draw the H diagonals AF, CH.

The angle ABF is made up of the angle ABC, together with the right angle CBF; the angle CBH is made up of the same angle ABC, together with the right angle ABH; hence the G 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 (Book I. Prop. V.).

F

E

The triangle ABF is half of the rectangle BE, because they have the same base BF, and the same altitude BD (Prop. II. Cor. 1.). 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 (Book I. Prop. III.); 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

B

D

K

square.

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 But the two rectangles BDEF, CDEG, taken together, make up the square BCGF: therefore the square BCGF, de scribed on the hypothenuse, is equivalent to the sum of the squares ABHL, ACIK, described on the two other sides; i other words, BC2=AB2+AC?.

3 Cor. 1. Hence the square of one of the sides of a right angled triangle is equivalent to the square of the hypothenuse diminished by the square of the other side; which is tas expressed: AB2-BC-AC2.

Cor. 2. It has just been shown that the square AH is equivalent to the rectangle BDEF; but by reason of the common altitude BF, the square BCGF is to the rectangle BDEF as the base BC is to the base BD; therefore we have

BC2: AB2: : BC: BD.

Hence the square of the hypothenuse is to the square of one of the sides about the right angle, as the hypothenuse is to the segment adjacent to that side. The word segment here denotes that part of the hypothenuse, which is cut off by the perpendicular let fall from the right angle: thus BD is the segment adjacent to the side AB; and DC is the segment adjacent to the side AC. We might have, in like manner,

BC2: AC2 :: BC: CD.

Cor. 3. The rectangles BDEF, DCGE, having likewise the same altitude, are to each other as their bases BD, CD. But these rectangles are equivalent to the squares AH, AI; therefore we have AB2: AC2 :: BD : DC.

Hence the squares of the two sides containing the right angle, are to each other as the segments of the hypothenuse which lie adjacent to those sides.

Cor. 4. Let ABCD be a square, and AC its H
diagonal: the triangle ABC being right an-
gled and isosceles, we shall have AC2=AB2+
BC2=2AB2: hence the square described on the A
diagonal AC, is double of the square described
on the side AB.

Ꮐ

C

E B F

This property may be exhibited more plainly, by drawing parallels to BD, through the points A and C, and parallels to AC, through the points B and D. A new square EFGH will thus be formed, equal to the square of AC. Now EFGH evidently contains eight triangles each equal to ABE; and ABCD contains four such triangles: hence EFGH is double of ABCD.

Since we have AC: AB221; by extracting the square roots, we shall have AC: AB :: √2:1; hence, the diagonal of a square is incommensurable with its side; a property which will be explained more fully in another place.

PROPOSITION XII. THEOREM.

In every triangle, the square of a side opposite an acute angle is than the sum of the squares of the other two sides, by twice the rectangle contained by the base and the distance from the acute angle to the foot of the perpendicular let fall from the opposite angle on the base, or on the base produced.

Let ABC be a triangle, and AD perpendicular to the base CB; then will AB2=AČ2+BC2—2BC × CD.

There are two cases.

First. When the perpendicular falls within the triangle ABC, we have BD=BC—CD, and consequently BD2-BC2+CD2—2BC × CD (Prop. IX.). Adding AD to each, and observing that the right angled triangles ABD, ADC, give AD+BD2 = AB2, and AD2+CD2=AC2, we have AB2=BC2+ AC2-2BC × CD.

Secondly. When the perpendicular AD falls without the triangle ABC, we have BD =CD-BC; and consequently BD2=CD2+ BC2-2CDX BC (Prop. IX.). Adding AD to both, we find, as before, AB2=BC2+AC2 -2BC x CD.

B

A

D

PROPOSITION XIII. THEOREM.

The perpendicular cannot fall within the triangle; for, if it fell at any point such as E, there would be in the triangle ACE, the right angle E, and the obtuse angle C, which is impossible (Book I. Prop. XXV. Cor. 3.): D C

с

In every obtuse angled triangle, the square of the side opposite the obtuse angle is greater than the sum of the squares of the other two sides by twice the rectangle contained by the base and the distance from the obtuse angle to the foot of the perpendicular let fall from the opposite angle on the base produced.

Let ACB be a triangle, C the obtuse angle, and AD perpendicular to BC produced; then will AB2=AC2+BC2+2BC × CD.