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

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.).

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 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 AI. 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; in other words, BC2=AB2+AC2.

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 taas expressed: AB2BC2_AC2.

Cor. 2. It has just been shown that the square AH is equi valent 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,

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 angled and isosceles, we shall have AC2=AB2+ BC2=2AB2: hence the square described on the diagonal AC, is double of the square described on the side AB.

A

D G

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 AC2: AB2 : 21; 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 less 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=AC2+ BC2-2BC x 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 AD2 + BD2= AB2, and AD2+CD2 AC, 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.

PROPOSITION XIII. THEOREM.

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 full 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.

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

B

X

hence the perpendicular falls without; and we have BD=BC +CD. From this there results BD2-BC2+CD2+2BC × CD (Prop. VIII.). Adding AD2 to both, and reducing the sums as in the last theorem, we find AB2=BC2+AC2+2BC × CD.

Scholium. The right angled triangle is the only one in which the squares described on the two sides are together equivalent to the square described on the third; for if the angle contained by the two sides is acute, the sum of their squares will be greater than the square of the opposite side; if obtuse, it will be less.

PROPOSITION XIV. THEOREM.

In any triangle, if a straight line be drawn from the vertex to the middle of the base, twice the square of this line, together with twice the square of half the base, is equivalent to the sum of the squares of the other two sides of the triangle.

Let ABC be any triangle, and AE a line drawn to the middle of the base BC; then will

2AE2+2BE2-AB2+AC2.

On BC, let fall the perpendicular AD

Then, by Prop. XII.

AC2-AE+EC2-2EC × ED.

And by Prop. XIII.

AB2-AE+EB2+2EB × ED.

B

ED

C

Hence, by adding, and observing that EB and EC are equal, we have

AB2+AC2-2AE2+2EB2.

Cor. Hence, in every parallelogram the squares of the sides

are together equivalent to the squares of the diagonals.

For the diagonals AC, BD, bisect each B other (Book I. Prop. XXXI.); consequently

the triangle ABC gives

AB+BC-2AE2+2BE2.

The triangle ADC gives, in like manner.

AD2+DC2-2AE2 + 2DE2.

E

D

Adding the corresponding members together, and observing that BE and DE are equal, we shall have

AB2 + AD2 + DC2 + BC2=4AE2+4DE2.

But 4AE2 is the square of 2AE, or of AC; 4DE2 is the square of BD (Prop. VIII. Cor.): hence the squares of the sides are together equivalent to the squares of the diagonals.

PROPOSITION XV. THEOREM.

If a line be drawn parallel to the base of a triangle, it will divide the other sides proportionally.

Let ABC be a triangle, and DE a straight line drawn par allel to the base BC; then wil!

AD : DB:: AE : EC.

Draw BE and DC. The two triangles BDE, DEC having the same base DE, and the same altitude, since both their vertices lie in a line parallel to the base, are equivalent (Prop. II. Cor. 2.).

The triangles ADE, BDE, whose common vertex is E, have the same altitude, and are to each other as their bases (Prop. VI. Cor.) ; hence we have

ADE: BDE :: AD: DB.

The triangles ADE, DEC, whose common vertex is D, have also the same altitude, and are to each other as their bases; hence

ADE: DEC:: AE: EC.

But the triangles BDE, DEC, are equivalent; and therefore, we have (Book II. Prop. IV. Cor.)

AD: DB:: AE: EC.

Cor. 1. Hence, by composition, we have AD+DB : AD :: AE+EC: AE, or AB : AD: AC: AE; and also AB : BD AC: CE.

Cor. 2. If between two straight lines AB, CD, any number of parallels AC, EF, GH, BD, &c. be drawn, those straight lines will be cut proportionally, and we shall have AE: CF; EG: FH: GB : HD.

For, let O be the point where AB and CD meet. In the triangle OEF, the line AC being drawn parallel to the base EF, we shall have OE : AE :: OF: CF, or OE: OF :: AE CF. In the triangle OGH, we shall likewise have OE : ÉG :: OF: FH, or OE: OF :: EG: FH. And by reason of the common ratio OE : OF, those two proportions give AE: CF B :: EG : FH. It may be proved in the

G

E

A

C

F

same manner, that EG: FH :: GB : HD, and so on; hence the lines AB, CD, are cut proportionally by the parallels AC, EF, GH, &c.