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PROPOSITION 45. PROBLEM. To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle.

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Let ABCD be the given rectilineal figure, and E the given angle.

It is required to describe a parallelogram equal to ABCD, and having an angle equal to E. Suppose the given rectilineal figure to be a quadrilateral.

Construction,

Join BD. Describe the parallelogram FH equal to the triangle ABD,

and having the angle FKH equal to the angle Ē. I. 42. To GH apply the parallelogram GM, equal to the triangle DBC, and having the angle GHM equal to E.

I. 44. Then shall FKML be the parallelogram required.

Proof.

Because each of the angles GHM, FKH = the angle E;

.. the angle FKH = the angle GHM.

To each of these equals add the angle GHK; then the angles FKH, GHK together = the angles GHM, GHK.

But since FK, GH are parallel, and KH meets them; :, the angles FKH, GHK together = two right angles; 1. 29. .:. also the angles GHM, GHK together=two right angles; .. KH, HM are in the same straight line.

I. 14. I. 14.

Again, because KM, FG are parallel, and HG meets them, .. the angle MHG= the alternate angle HGF.

I. 29. To each of these equals add the angle HGL; then the angles MHG, HGL together= the angles HGF, HGL.

But because HM, GL are parallel, and HG meets them, .. the angles MHG, HGL together=two right angles : 1. 29. ... also the angles HGF, HGL together=two right angles :

.:. FG, GL are in the same straight line. And because KF and ML are each parallel to HG, Constr. therefore KF is parallel to ML;

I. 30. and KM, FL are parallel ;

Constr. .:. FKML is a parallelogram. Def. 36. Again, because the parallelogram FH= the triangle ABD,

and the parallelogram GM = the triangle DBC; Constr. .. the whole parallelogram FKML = the whole figure ABCD;

and it has the angle FKM equal to the angle E. By a series of similar steps, a parallelogram may be constructed equal to a rectilineal figure of more than four sides.

Q.E.F.

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The following Problem is important, and furnishes a useful application of the principles of the foregoing propositions.

ADDITIONAL PROBLEM.

To describe a triangle equal in area to a given quadrilateral.

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Let ABCD be the given quadrilateral.

It is required to describe a triangle equal to ABCD in area. Construction.

Join BD. Through C draw CX parallel to BD, meeting AD produced in X.

Join BX. Then XAB shall be the required triangle. Proof. Now the triangles XDB, CDB are on the same base DB and between the same parallels DB, XC; .. the triangle XDB=the triangle CDB in area.

I. 37. To each of these equals add the triangle ADB ;

then the triangle XAB=the figure ÅBCD.

EXERCISE.

Construct a rectilineal figure equal to a given rectilineal figure, and having fewer sides by one than the given figure.

Hence shew how construct a triangle equal to a given recti. lineal figure

PROPOSITION 46. PROBLEM.

To describe a square on a given straight line.

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I. 3.

Let AB be the given straight line.

It is required to describe a square on AB.
Constr. From A draw AC at right angles to AB;

1. 11. and make AD equal to AB.

Through D draw DE parallel to AB; I. 31. and through B draw BE parallel to AD, meeting DE in E.

Then shall ADEB be a square. Proof. For, by construction, ADEB is a parallelogram: . . AB=DE, and AD=BE.

I. 34. But AD= AB ;

Constr. .:: the four straight lines AB, AD, DE, EB are all equal ;

that is, the figure ADEB is equilateral. Again, since AB, DE are parallel, and AD meets them, .. the angles BAD, ADE together=two right angles ; 1. 29.

but the angle BAD is a right angle ; Constr.

... also the angle ADE is a right angle. And the opposite angles of a parallelogram are equal ; I. 34. .. each of the angles DEB, EBA is a right angle:

that is the figure ADEB is rectangular. Hence it is a square, and it is described on AB.

Q.E.F. COROLLARY. If one angle of a parallelogram is a right angle, all its angles are right angles.

PROPOSITION 47. THEOREM.

In a right-angled triangle the square described on the hypotenuse is equal to the sum of the squares described on the other two sides.

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Let ABC be a right-angled triangle, having the angle BAC a right angle.

Then shall the square described on the hypotenuse BC be cqual to the sum of the squares described on BA, AC. Construction. On BC describe the square BDEC; I. 46. and on BA, AC describe the squares BAGF, ACKH. Through A draw AL parallel to BD or CE;

1. 31. and join AD, FC. Proof. Then because each of the angles BAC, BAG is a right angle, ... CA and AG are in the same straight line.

1. 14.
Now the angle CBD=the angle FBA,
for each of them is a right angle.

Add to each the angle ABC:
then the whole angle ABD = the whole angle FBC.

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