311. COROLLARY.

The sum of the squares on the four sides of a parallelogram is equivalent to the sum of the squares on the diagonals, because the diagonals of a parallelogram bisect each other.

GIVEN, any polygon N.

REQUIRED, to describe a square equivalent to N.

CONSTRUCTION. Describe, by 263, the rectangle ABCD = N.
Then, if AB = BC, the required square is ABCD.

If AB be not equal to BC, produce BA, and cut off AF = Bisect BF in G, and, with center G and radius GB, describe FLB. Produce DA to meet the circle in H.

The square on AH shall be equal to N.

AD.

PROOF. Join GH. Then, by 296, because the sect BF is divided equally in G and unequally in A,

.. rectangle BA,AF + AG2

GF2

=

GH2
GĦ2 = AH2 + AG2,

by 243;

:. AĦ2 rectangle BA,AF = rectangle BA,AD = N.

PROBLEM II.

313. To divide a given sect into two parts so that the rectangle contained by the whole and one of the parts shall be equal to the square on the other part.

REQUIRED, to find a point C such that rectangle AB,BC = AC2.
CONSTRUCTION. On AB, by 241, describe the square ABDF.
By 136, bisect AF in G. Join BG.

Produce FA, and make GH = GB.
On AH describe the square AHKC.

PROOF. Produce KC to L. Then, by 298, because FA is bisected in G and produced to H, rectangle FH,HA + AG

GH2

GB2

PROBLEM III.

314. To describe an isosceles triangle having each of the angles at the base double the angle at the vertex.

CONSTRUCTION. Take any sect, AB, and, by 313, divide it in C so that rectangle AB,AC = BC'. With center C and radius CB, describe a circle.

With same radius, but center A, describe a circle intersecting the preceding circle in D. Join AD, BD, CD.

ABD will be the triangle required.

PROOF. By construction, CB = CD = DA.

By 134, bisect & ADC by DF. By 137, DF is 1 AC, and bisects AC.

Then

AB2 + AD2 = BD2 + 2AB,AF.

(306. The square on a side opposite an acute angle is less than the squares on the other two sides by twice the rectangle of either and the projection of the other on it.)

..

by our initial construction, BC2 = 2AB,AF.
:. AB2 + AD2 = BD2 + BC.

But, by construction, AD = BC, :: AB2 = BD2, :. AB = BD,

...

= 2 X B.

× ADB = ¥ DAB = × ACD = 4 B + ¥ BDC = (173. The exterior angle of a triangle equals the sum of the two opposite interior angles.)

BOOK III.

THE CIRCLE.

I. Primary Properties.

315. If a sect turns about one of its end points, the other end point describes a curve called the Circle.

316. The fixed end point is called the Center of the circle. 317. The moving sect in any position is called a Radius of the circle.

318. As the motion of a sect does not enlarge or diminish it, all radii are equal.

319. Since the moving sect, after revolving through a perigon, returns to its original position, therefore the moving end point describes a closed curve.

This divides the plane into two surfaces, one of which is swept over by the moving sect. This finite plane surface is

called the surface of the circle.

Any part of the circle is called an Arc.