Page images
PDF
EPUB

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 thus expressed: AB2BC?—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 angled and isosceles, we shall have AC2=AB2+ BC2-2AB: hence the square described on the diagonal AC, is double of the square described on the side AB.

E B F

G

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: 2:1; 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 AD2 to each, and observing that the right angled triangles ABD, ADC, give AD2+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, ÁB2=BC2+AC2 -2BC × CD.

[blocks in formation]

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

hence the perpendicular falls without; and we have BD=BC +CD. From this there results BD2-BC2+CD2+2BC × CD (Prop. VIII.). Adding AD 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.

ACAE+EC-2EC x ED.

And by Prop. XIII.

X

AB2-AE+EB2+2EB × ED.

B

ED

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

AB2+ BC-2AE2+2BE2.

The triangle ADC gives, in like manner.

AD2+DC2-2AE2+2DE2.

E

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

AB2+AD2+DC+BC2-4AE2+4DE.

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,

11

PROPOSITION XV. THEOREM.

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

Let ABC bé a triangle, and DE a straight line drawn parallel to the base BC; then will.

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.

[blocks in formation]

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
:: OF: FH, or OE: OF EG: FH.
And by reason of the common ratio OE
OF, those two proportions give AE CF B

:

EG

:: EG : FH. It may be proved in the

G

[ocr errors]

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.

[ocr errors]

PROPOSITION XVI. THEOREM.

Conversely, if two sides of a triangle are cut proportionally by a straight line, this straight line will be parallel to the third side.

A

In the triangle ABC, let the line DE be drawn, making AD: DB AE EC: then will DE be parallel to BC. For, if DE is not parallel to BC, draw DO parallel to it. Then, by the preceding theorem, we shall have AD DB:: AO: OC. But by hypothesis, we have AD : DB :: AE: EC: hence we must have AO: OC :: AE: EC, or AO: AE :: OC: EC; an impossible result, since AO, the one antecedent, is less than its consequent AE, and OC, the other antecedent, is greater than its consequent EC. Hence the parallel to BC, drawn from the point D, cannot differ from DE; hence DE is that parallel.

D

B

Scholium. The same conclusion would be true, if the proportion AB AD: AC AE were the proposed one. For this proportion would give AB-AD : AD : : AC—AE : AE, or BD AD :: CĚ : AE.

PROPOSITION XVII. THEOREM.

The line which bisects the vertical angle of a triangle, divides the base into two segments, which are proportional to the adjacent sides.

In the triangle ACB, let AD be drawn, bisecting the angle CAB; then will

BD CD: AB: AC.

Through the point C, draw CE E parallel to AD till it meets BA produced.

In the triangle BCE, the line AD is parallel to the base CE; hence we have the proportion (Prop. XV.),

D

BD: DC :: AB : AE. But the triangle ACE is isosceles: for, since AD, CE are parallel, we have the angle ACE DAC, and the angle AEC-BAD (Book I. Prop. XX. Cor. 2 & 3.); but, by hypothesis, DAC BAD; hence the angle ACE AEC, and consequently AE-AC (Book I. Prop. XII.). In place of AE in the above proportion, substitute AC, and we shall have BD DC: AB: AC.

:

« PreviousContinue »