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square BCGF, described on the hypothenuse, is equivalent to the sum of the squares ABHL, ACIK, described on the other two sides; in other words, BC=AB+ AC. Therefore,

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

Cor. 1. Hence, by transposition, 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, AB-BC-AC2.

Cor. 2. It has just been shown that the square AH is equal 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.

Note. The word segment here denotes that part of the hypothenuse, which is cut off by the perpendicular let fall from the rightangle 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 AB, AC2; 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.

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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 a triangle ABC, if the angle C is acute, the square of the opposite side will be less than the sum of the squares of the sides which contain the angle C, by twice the rectangle BC × CD, the line AD being perpendicular to BC. That is, AB-AC+BC-2(BC x CD).

First. When the perpendicular falls within the triangle ABC, we have BD=BC-CD, and consequently (9. 4,) BD2 =BC2+CD2-2(BC x CD). Adding AD' to each, and observing that the right-angled triangles ABD, ADC give AD3 +BD2 AB, and AD2+CD2=AC', we have AB2=BC2+ AC2-2(BC × CD).

Second. When the perpendicular AD falls without

the triangle ABC, we have

BD=CD-BC; and consequently BD'=CD2+BC2

-2 (CD × BC).

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Adding

CD

B

AD' to both, we find, as before, AB2=BC+AC2=2(BC× CD). Hence,

In every triangle the square of the side subtending either of the acute angles, is less than the sum of the squares of the other two sides, by twice the rectangle contained by either of these sides and the straight line intercepted between the perpendicular let fall upon it from the opposite angle, and the acute angle.

THEOREM.
PROPOSITION XIII.

In a triangle ABC, if the angle C is obtuse, the square of the opposite side AB, will be greater than the sum of the squares of the sides, which contain the angle C, by twice the rectangle BCX CD, AD being perpendicular to BC. That is,

AB2=AC2+BC2+2(BC× CD).

A

The perpendicular cannot fall within the triangle; for if it fell at any point E, the triangle ACE would have both the right-angle E, and the obtuse angle C; which is impossible; D (28. 1. Cor. 3 ;) hence the perpendicu

CE

lar falls without; and we have BD=BC+CD.

From this

there results, (8. 4,) BD'=BC+CD2+2(BC × CD). Adding AD' to both, and reducing the sums as in the last theorem, we find AB2=BC2+AC2+2(BC x CD). Hence,

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In every obtuse-angled triangle, the square of the side sul tending the obtuse-angle, is greater than the sum of the square

of the other two sides by twice the rectangle contained by that side upon which, when produced, the perpendicular falls, and the straight line intercepted between the perpendicular and the obtuse angle.

Scholium. The right-angled triangle is the only one in which the sum of the squares of two sides is equal to the square of the third side; 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.

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In any triangle ABC, if a straight line AE be drawn from the vertex to the middle of the base BC, the sum of the squares of the other two sides AB, AC, will be equal to twice the square of half the base plus twice the square AE. That is,

AB2+AC-2AE2+2BE2.

On BC, let fall the perpendicular AD. Then will AC2=AE2+ EC2-2EC × ED. (12. 4.) And AB2 AE2+EB2+2EB×ED.

Hence by adding the corresponding sides together, and observing that

EB and EC are equal, we have

B

ED

AB2+AC-2AE2+2EB2. Therefore,

If a straight line be drawn from one of the angles of any triangle to the middle of the opposite side or base, the sum of the squares of the other two sides, will be equal to twice the square of half the base plus twice the square of this line.

Cor. In every parallelogram the squares of the sides are together equal to the squares of the diagonals.

For the diagonals AC, BD bisect B each other; (33. 1;) consequently, AB2+BC2=2AE2+2BE2. In like

manner, AD2+DC=2AE+2DE2. Adding the corresponding members

A

E

together, and observing that BE and DE are equal, we shall have

AB2+AD3+DC2+BC2-4AE2+4DE2.

But 4AE2 is the square of 2AE, or of AC; (8. 4. Sch. 2;) 4DE2 is the square of BD: hence the squares of the sides are together equal to the squares of the diagonals.

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The line DE, drawn parallel to the base of the triangle ABC, divides the sides AB, BC proportionally; that is,

AD: DB:: AE: EC.

Join 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. (2. 4. Cor. 2.)

The triangles ADE, BDE whose common vertex is E, having the same altitude, and are to each other as their bases: (6. 4. Cor.:) hence we have

B

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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.

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