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PROPOSITION VIII. THEOREM 355. The square constructed on the sum of two lines is equivalent to the sum of the two squares constructed on these lines, increased by twice the rectangle of these lines.

H

K

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Нур.

Line AC= AB + BC. То prove

the square on AC = to the sum of the squares on
AB and BC increased by twice the rectangle contained by
AB and BC.
Proof. On AC, construct the square

ACKG.
On AG, lay off AD= AB.
Through B, draw BH || AG, through D, draw DF|AC.

The square AK is thus divided into four parts, AE, EK, GE, and BF, all of which are rectangles.

(Why?) But AE is the square on AB;

(Why?) EK is the square = to the square on BC;

(Why?) GE is = to the rectangle contained by AB and BC; (Why ?) and BF is = to the rectangle contained by AB and BC. (Why ?)

.. the square on AC is to the sum of the squares on AB and BC, increased by twice the rectangle contained by AB and BC.

356. SCHOLIUM. This theorem illustrates geometrically the algebraic formula,

(a + b = a + 2 ab + b?.

Q.E.D.

PROPOSITION IX. THEOREM 357. The square constructed on the difference of two lines is equivalent to the sum of the two squares constructed on these lines, diminished by twice the rectangle of these lines.

d

Hint. - Demonstrate that d2 is obtained when we construct geometrically

a2 ab + b2 ab.

358. SCHOLIUM 1. This theorem illustrates geometrically the algebraic formula,

(a - b) = a? - 2 ab + b?.

359. Scholium 2. Any homogeneous equation of the second degree can be illustrated geometrically.

ac.

Ex. 806. Prove, geometrically, the algebraic formula,

a(b + c)= ab + ac. Ex. 807. Prove, geometrically, the algebraic formula,

a(b c)= ab Ex. 808. Prove, geometrically, the algebraic formula,

(a + b)(a b)= a– b2. Ex. 809. Prove, geometrically, the algebraic formula,

(a + b)2-(a - b)2= 4 ab. Ex. 810. Prove, geometrically, the algebraic formula,

(a + b)(c + d)= ac + bc + ad + bd.

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Given. Rectangle ABCD.
Required. A square - to ABCD.

Construction. Construct a mean proportional AF between
AB and AD. The square on AF is the required square.
Proof.
AB: AF- AF: AD.

(Con.) .:. AB X AD=

(262) area ABCD= area AG.

AF,

or

Q.E.F.

361. SCHOLIUM. The mean proportional may be constructed by any of the methods of Book III. The two indicated in the diagram, however, are the most useful ones.

Ex. 811. To transform a parallelogram into a square.
Ex. 812. To transform a triangle into a parallelogram.
Ex. 813. To transform a triangle into a square.
Ex. 814. To transform a pentagon into a square.

Ex. 815. To construct a square equivalent to one-third of a given triangle.

PROPOSITION XI. PROBLEM 362. To construct a square that shall be any given part of a given square.

H

G

B В

square, AH.

Given. Square AC.
Required. A square equivalent to 4 of AC.
Construction. On AD, lay off AE = 4 AD.
Draw EFI AD, and transform rectangle AEFB into a

(360)
AH is the required square.
Proof.
AH = AF.

(Con.) AF = 4 AC.

(335) .. AH = 4 AC.

Q.E.F.

m

n

363. SCHOLIUM. If, instead of 4, the ratio

is given, where m and n are two given lines, make AE the fourth proportional to n, m, and AD.

Ex. 816. To construct a square three times as large as a given square.

Ex. 817. In a given square ABCD, to construct four squares, all containing the angle A, and having areas respectively equal to }, }, $, and he of ABCD.

N

Ex. 818. In the diagram for Prop. XI, find two lines whose squares have the ratio AE : ED.

Ex. 819. In the same diagram, find a rectangle equivalent to the square on GD.

Ex. 820. In the same diagram, find a square equivalent to the sum of the squares on AG and GD.

PROPOSITION XII. THEOREM

364. In a right triangle the sum of the squares of the arms is equivalent to the square of the hypotenuse.

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Hyp. AH and BF are squares on the arms, AD the square on the hypotenuse of the right triangle ABC.

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Proof. From B draw BL | AE, intersecting AC and ED in I and L, respectively. Draw BE and KC.

< HBC = st. 2,

or

HBC is a straight line.

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