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It is required to apply to AB a parallelogram equal to C, and

having an angle equal to D. Make the parallelogram BEFG equal to 0, and having the angle EBG equal to D,

(I. 42) so that BE may be in the same straight line with AB; produce FG to H, through A draw AH parallel to EF,

(I. 31) and join HB.

F E

K

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Then, because the straight line HF meets the parallels AH,

EF, therefore the angles AHF, HFE are together equal to two right angles ;

(I. 29) wherefore the angles BHF, HFE are together less than two right

angles; therefore HB, FE will meet if produced.

(Ax. 12) Let them be produced and meet in K, through K draw KL parallel to FH, and produce HA, GB to meet LK in L, M.

Then BL shall be the parallelogram required.
Now HLKF is a parallelogram,

(Constr.) of which the diameter is HK; also AG, ME are the parallelograms about HK, and LB, BF are their complements; therefore the parallelograms LB, BF are equal. (I. 43) But BF is equal to the triangle C;

(Constr.) wherefore LB is equal to C.

Again, because the angle EBG is equal to ABM, (I. 15) and likewise to the angle D;

(Constr.) therefore the angle ABM is also equal to D.

(Ax. 1)

Wherefore to the straight line AB the parallelogram LB

has been applied, equal to the triangle C, and having the angle ABM equal to the angle D.

Q. E. F.

97. EXERCISE LII.

*

1. State the proposition which may be thus enunciated :

“Construct a parallelogram, given one side, one angle, and a

triangle of equal area.” 2. Show how to make a rectangle, having given one side, and also a triangle of equal area.

3. Prove any two adjacent angles of a parallelogram are together equal to two right angles. Give examples from the figure of XLV., in which FH, HL are each a parallelogram.

(Questions 4-9 refer to the same figure.)

4. Given also that the angles K and L are each equal to E, show that GHM is equal to K.

5. By adding KHG to each equal, prove the adjacent angles at H are equal to two right angles.

6. Draw the inference from this, giving the reference.

7. Prove that the angles FGH and I are equal; given, as before, that the angle E is equal to each of the angles K, L.

8. Add HGL to each of these equals, and then proceed to prove FGL one straight line.

9. If FH, HL are parallelograms, prove FK, LM are parallels.

10. If ABC be any triangle, having BC for base, and D is taken bisecting AB, and then joined to C, show that the triangle ABC is double of the triangle DBC.

11. Given ABC any triangle and BC its base; apply to BC a parallelogram equal to ABC, and having an angle equal to a given rectilineal angle, without making use of Prop. XLIII.

12. Given a triangle; apply to any side of it a rectangle equal in area to the triangle, without applying XLIII.

98, PROPOSITION XLV.-PROBLEM. To describe a parallelogram equal to a given rectilineal figure,

and having an angle equal to a given rectilineal angle. Let ABCD be the given rectilineal figure, and E the given rectilineal angle.

I

It is required to describe a parallelogram, that shall be equal to ABCD, and have an angle equal to E.

Join DB. Describe the parallelogram FH equal to the triangle ADB, and having the angle FKH equal to E;

(I. 42) to the straight line GH apply the parallelogram GM equal to

DBC, and having the angle GHM equal to E. (I. 44) Then the figure FKML shall be the parallelogram required.

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Because E is equal to each of the angles FKH, GHM, therefore FKH is equal to GHM; add to each the angle KHG; therefore the angles FKH, KHG are equal to KHG, GHM; but FKH, KHG are equal to two right angles ;

(I. 29) therefore, also, KHG, GHM are equal to two right angles, that is, at the point H in the straight line GH, the two KH,

HM, upon the opposite sides of it, make the adjacent

angles equal to two right angles ; therefore HK is in the same straight line with HM. (I. 14)

And because HG meets the parallels KM, FG, the angle MHG is equal to the alternate angle HGF; (I. 29) add to each the angle HGL; therefore MHG, HGL are equal to HGF, HGL; but MHG, HGL are equal to two right angles ; (1. 29) therefore, also, HGF, HGL are equal to two right angles;

wherefore FG is in the same straight line with GL. (I. 14)

Again, because KF is parallel to HG, and HG to ML, therefore KF is parallel to ML;

(I. 30) and KM has been proved parallel to FL;

therefore the figure FKML is a parallelogram.

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And since FH is equal to the triangle ABD,

and GM to the triangle BDC;

therefore the whole FKML is equal to the whole ABCD. Wherefore a parallelogram FKML has been described equal to the given rectilineal figure ABCD, and having the angle FKM equal to the given angle E.

Q. E. F.

COR. From this it is manifest how, to a given straight line, to apply a parallelogram, which shall have an angle equal to a given angle, and shall be equal to a given rectilineal figure of any number of sides;

namely, by applying to the given straight line a parallelogram equal to the first triangle ABD, and having an angle equal to the given angle (I. 44); and then applying to a side of this parallelogram another equal to the second triangle; and so on.

draw DH parallel to AC,

meeting BC produced in H;

99. REDUCTION OF A POLYGON TO A TRIANGLE OF EQUIVALENT AREA.

By means of Props. XLIV. and XLV., Euclid reduces any polygon to a parallelogram of equivalent area. It is considerably easier to reduce a polygon to a triangle of the same area; and, as the method is of practical utility, it will now be given.

PROBLEM.-To construct a triangle equal in area to a given rectilineal figure.

Let ABCD be the given figure.

It is required to construct a triangle equal to ABCD.
Join AC;

(I. 31)

and join AH.

Then ABH is the triangle required.
Because the triangles HAC, DAC are upon

the same base AC, and between the same
parallels DH, AC;

B

therefore HAC is equal to DAC.

To each equal add the triangle ABC;

then the whole triangle HAB is equal to the whole figure ABCD. Wherefore, a triangle ABH has been constructed equal in area to the given rectilineal figure ABCD. Q. E. F. Continuing the preceding, we may take a parallelogram equal to the tri

(I. 37)

angle ABH (I. 42); and the whole proposition would then be equivalent. to I. 45.

In the proof given, just as in Euclid, the rectilineal figure is taken quadrilateral. If there be another side to the figure, it must be divided into three triangles by drawing straight lines from one angle; and for every additional side there will be an additional triangle.

The process would then be to replace two adjacent triangles by one, as just shown; then replace this new one and an adjacent triangle in the same way; and so on till all are reduced.

100. EXERCISE LIII. 1. Make a figure to XLIII., having K the middle point of AC; then show that the triangles AHK, KFC are equal by I. 26.

2. Also prove that under the same circumstances the parallelograms about AC are equal.

3. Join HF, in the figure just referred to, and then prove HF parallel to AC.

4. Also join EG, and prove it parallel to HF.

5. In the usual figure to XLIII., join EG and HF; and prove the triangles EGK, FHK equal.

6. If FG be joined and the triangle KFG be added to each equal, what is the resulting pair of equals ?

7. Now join EH, and prove EH parallel to GF.

8. Let EH and GH cut AC in L, M respectively ; prove LM to be half of AC.

9. Given a triangle and a straight line, make a rectangle upon the straight line equal to the triangle.

10. Given a triangle and a straight line, make a rectangle upon the straight line double of the triangle,

11. Show how to make a parallelogram double of a given rectilineal figure, having an angle equal to two-thirds of a right angle.

12. Describe a rhombus equal to a given rectilineal figure.

SECTION XI. PROPOSITIONS XLVI –XLVIII. Squares and the sides of a right-angled Triangle.

101. PROPOSITION XLVI.-PROBLEM.

To describe a square upon a given straight line. Let AB be the given straight line. It is required to describe a square upon AB. From the point A draw AC at right angles to AB; (I. 11)

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