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PROBLEM 20. To inscribe a regular octagon in a given circle. PROBLEM 21. To inscribe a regular hexagon, or an equilateral triangle, or a regular dodecagon, in a given circle.

PROBLEM 22. To inscribe a regular decagon, or pentagon, in a given circle.

PROBLEM 23. To inscribe a regular pentedecagon in a given circle.

PROBLEM 24.

To circumscribe a square to a given circle.

Inscribe a square in the given circle, and, through each vertex, draw a tangent to the circle: the quadrangle thus formed is a square circumscribed to the circle.

For, it is regular [94], and its sides are tangent to the given circle. PROBLEM 25. To circumscribe an equilateral triangle, or a regular hexagon, octagon, decagon, dodecagon, or pentedecagon, to a given circle.

BOOK IV.

ON AREAS.

BOOK IV.

ON AREAS.

SECTION I.

ON EQUIVALENCE GENERALLY.

DEFINITIONS.

I. When two parallels contain opposite sides of a parallelogram, or the base and apex of a triangle, the parallelogram, or the triangle, is said to be between these parallels.

Parallelograms or triangles which are between the same parallels have obviously equal altitudes.

2. Of the four partial parallelograms into which a parallelogram is divided, by parallels drawn to two adjacent sides, through a point of one of the diagonals, those about the diagonal are said to be completed by the others; the latter are called the complements of the former.

THEOREM 1.

Two parallelograms having equal bases and altitudes, are equivalent to each other.

Let there be two parallelograms A B C D and A E F D, having equal bases and altitudes.

Let these parallelograms be applied to each other so that their

B

equal bases coincide with each other in A D, and let C E be joined; let also C H and E I be perpendiculars from the vertices C and E to

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the common base A D and its prolongation: then CH is the altitude of the parallelogram A C [III. Def. 20] and E I is the altitude of the parallelogram A F. Because the lines C H and E I are equal to each other [Hyp.], CE is parallel to AD [I. 20]; but BC and E F are also parallel to A D [Hyp.]: therefore B C E F is a straight line parallel to A D [I. 17 iv]. Now, the quadrangle A B F D is composed of the parallelogram A C and the triangle D C F, but it is also composed of the parallelogram A F and the triangle A B E. Because the sides of the angles BA E and C D F are parallel and have the same direction, the angle B A E is equal to the angle C D F [II. 23]; because also A B is equal to C D, and A E to D F [III. 29], the triangles D CF and AB E are equal to each other [III. 17]: thence if either of these two triangles be taken from the quadrangle A B F D, the remainders will be equal in magnitude [ii]; therefore the parallelograms A C and A F are equivalent to each other, W. W. T. B. D.COROLLARY. If two parallelograms having the same base or equal bases, be between the same parallels, they are equivalent to each other. (Eucl. I. 35 and 36.)

Scholium. The perimeters of two equivalent parallelograms having equal bases may differ infinitely.

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