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AC, the longest side of the one,
: AB, the longest side of the other,
:: AB, the shortest side of the one,

: AF, the shortest side of the other. Also in the similar A ABC and FBC,

AC, the longest side of the one,
: BC, the longest side of the other,
:: BC, the medium side of the one,

: FC, the medium side of the other.

Q. E. D.

335. COR. 1. The squares of the two legs of a right triangle are proportional to the adjacent segments of the hypotenuse. The proportions in II. give, by § 295,

AB ACX AF, and BC= ACX CF.

=

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336. COR. 2. The squares of the hypotenuse and either leg are proportional to the hypotenuse and adjacent segment.

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337. COR. 3. An angle inscribed in a semicircle is a right angle (§ 264). Therefore,

I. The perpendicular from any point in the circumference to the diameter of a circle is a mean proportional between the segments of the diameter.

A

D

B

II. The chord drawn from the point to either extremity of the diameter is a mean proportional between the diameter and the adjacent segment.

REMARK. The pairs of corresponding sides in similar triangles may be called longest, shortest, medium, to enable the beginner to see quickly these pairs; but he must not forget that two sides are homologous, not bogange thou poor to be the longest or the shortest sides but h

338. The sum of the squares of the two legs of triangle is equal to the square of the hypoten

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Let ABC be a right triangle with its right an

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By adding, AB + BC2 = AC(AF+CF) = AC2, 339. COR. The square of either leg of a right triang to the difference of the squares of the hypotenuse and the 340. SCHOLIUM. The ratio of the diagonal of a square to the side is the incommensurable number √2. For if AC is the diagonal of the square ABCD, then

AC2 = AB2 + BC2, or AC2 = 2AB2.

D

A

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Since the square root of 2 is incommensurable, the and side of a square are two incommensurable lines.

C

341. The projection of a line CD upon a straight li that part of the line AB comprised between the perpendiculars CP and DR let fall from the extremities of CD. Thus, PR is the projection of CD upon AB.

NUMERICAL PROPERTIES OF FIGURES.

159

PROPOSITION XVII. THEOREM.

342. In any triangle, the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides and the projection of the other upon that side.

A

2

AC2 = AB TRE - 2 CBR)

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Let C be an acute angle of the triangle ABC, and

DC the projection of AC upon BC.

To prove
Proof. If D fall upon the base (Fig. 1),

AB2 = BC2 + AC2 – 2 BC× DC.

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DB2 = BC2 + DC2 - 2 BCX DC.
Add AD to both sides of this equality, and we have
AD2 + DB2 = BC2+AD2+DC2 — 2 BC× DC.

But

and

AD + DB2 = AB2,

AD2 + DC2 = AC2,

§ 338

(the sum of the squares of the two legs of a rt. ▲ is equal to the square

of the hypotenuse).

Put AB and AC2 for their equals in the above equality,
AB2 = BC2 + AC2 – 2BC× DC.

Q. E. D.

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343. In any obtuse triangle, the square of opposite the obtuse angle is equal to the su squares of the other two sides increased by product of one of those sides and the proj the other upon that side.

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Let C be the obtuse angle of the triangle CD be the projection of AC upon BC produce

To prove Proof.

AB2 = BC2 + AC2+2BC× DC.

=

DB=BC+ DC.

Squaring, DB BC2 + DC2 + 2 BC× DC.
Add AD' to both sides, and we have

But

and

AD2 + DB2 = BC2+AD2+DC2+2 BC ×

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(the sum of the squares of the two legs of a rt. ▲ is equal to

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of the hypotenuse).

for their equals in the above

BC2+ AC+2 BCX DC.

NOTE. The last three theorems enable us to compute t

the altitudes if the lengths of the three sides of a triangle ar

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344. I. The sum of the squares of two sides of a triangle is equal to twice the square of half the third side increased by twice the square of the median upon that side.

II. The difference of the squares of two sides of a triangle is equal to twice the product of the third side by the projection of the median upon that side.

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In the triangle ABC let AM be the median, and MD the projection of AM upon the side BC. Also let AB be greater than AC.

To prove

I. AB2 + AC2 = 2BM2+2AM2.

II. AB2 - AC2 = 2 BC× MD.

Proof. Since AB> AC, the ≤ AMB will be obtuse, and the AMC will be acute.

Then AB BM2+ AM2+2 BM× MD,

§ 158

§ 343

(in any obtuse ▲ the square of the side opposite the obtuse 4 is equal to the sum of the squares of the other two sides increased by twice the product of one of those sides and the projection of the other on that side);

=

2

AC2 - MC2 + AM2 — 2 MC × MD,

and
$ 342
(in any ▲ the square of the side opposite an acute is equal to the sum of
the squares of the other two sides diminished by twice the product of one
of those sides and the projection of the other upon that side).
Add these two equalities, and observe that BM= MC.
Then AB2+ AC2 = 2BM2+2AM2.

Subtract the second equality from the first.

Then

AB-AC 2 BCX MD.

Q. E. D.

NOTE. This theorem enables us to compute the lengths of the medians

if the lengths of the three sides. fthe trienal

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