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To make a parallelogram that shall have its opposite sides equal to two given right lines, and one of its angles equal to a given rectilineal angle.
Let AB and c be two given right lines, and d a given reclilineal angle; it is required to make a parallelogram that shall have its opposite sides equal to AB and c, and one of its angles equal to D.
At the point A, in the line AB, make the angle BAF equal to the angle D (Prop. 20.) and the fide AF equal to c (Prop. 3.)
Also, make fe parallel and equal to AB (Prop. 28 and 3.), and join Be; then will AE be the parallelogram required.
For, since fe is parallel and equal to AB (by Conf.), "BE will be parallel and equal to AF (Prop. 30.); whence the figure ae is a parallelogram.
And, because AF is equal to c (by Canft.) BE will also be equal to c; and the angle BAF was made equal to the angle D.
The opposite sides of the parallelogram AE are, therefore, equal to the two given lines AB and c; and one of its angles is equal to the given angle D, as was to be done.
PRO P. XXXIV. THEOREM.
If two sides of a triangle be bisected, the right line joining the points of bisection, will be parallel to the base, and equal to one half of it.
Let ABC be a triangle, whose fides CA, CB are bisected in the points D, É ; then will the right line De, joining those points, be parallel to AB, and equal to one half of it.
For, in de produced, take er equal to ED (Prop. 3-), and join BF:
Then, fince Ec is equal to EB (by Hyp.), ED 'to EF (by Conft.) and the angle dec to the angle ber (Prop. 15.), the fide BF will also be equal to the side Dc, or its equal -DA, and the angle EFB to the angle EDC (Prop. 4.)
And, because the right line DF intersects the two right lines CD, FB, and makes the angle EDC equal to the alternate angle EFB, BF will be parallel to Dc or DA (Prop. 24.)
The right lines BF, AD, therefore, being equal and parallel, the lines DF, AB, joining their extremes, will also be equal and parallel (Prop. 30.)
But Dr is the double of DE (by Const.); consequently AB is also the double of de; that is de is the half of AB, 9
Q. E. D.
PRO P. XXXV. PROBLEM.
To divide a given finite right line into any proposed number of equal parts.
Let AB be the given right line ; it is required to divide it into a certain proposed number of equal parts.
From the point A, draw any right line Ac, in which take the equal parts AD, DE, EC, at pleasure, (Prop. 1.) to the number proposed.
Join BC; and parallel thereto draw the right lines EF, DG, (Prop. 28.) cutting AB, in F and G; then will AB be divided into the same number of equal parts with AC, as was required.
For take EH, CK, each equal to DG (Prop. 3.), and join D, H and E, K.
Then, since DG is parallel to EF (by Const.), and AE intersects them, the outward angle ADG will be equal to to inward opposite angle DEH (Prop. 25.)
And, because the sides AD, DG of the triangle AGD, are equal to the sides DE, Eh of the triangle DHE (by Conft.), and the angle AdG is equal to the angle DEH, the base AG will also be equal to the base dh, and the angle DAG to the angle EDH (Prop. 4.)
But, since the right line ae intersects the two right lines DG, EF, and makes the outward angle Edh equal
to the inward opposite angle DAG, DH will be parallel to AG or GF (Prop. 23.)
And, in the same manner it may be shown that Ek is also equal to AG, and parallel to AG or FB.
The figures GH, FK, therefore, being parallelograms, the fide DH will be equal to the side GF, and the side ek to the side FB (Prop. 31.)
But du, ek have been each proved to be equal to AG; consequently GF, FB are, also, each equal to AG; whence the line AB is divided into the same number of equal parts with ac, as was to be done.
1. A rectangle is a parallelogram whose angles are all right angles.
2. A fquare is a rectangle, whose sides are all equal to each other.
3. Every rectangle is said to be contained by any two of the right lines which contain one of the right angles.
4. If two right lines be drawn through any point in the diagonal of a parallelogram, parallel to its opposite fides, the figures which are intersected by the diagonal are called parallelograms about the diagonal.
5. And the other two parallelograms, which are not intersected by the diagonal, are called complements to the parallelograms which are about the diagonal.