SUMMARY OF PROPOSITION XLIII. THEOREM.

The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another. Cons.-Nil.

The figure is divided into three equal pairs of triangles (I. 34). One of each of the smaller triangles + one complement one large triangle.

D

A

=

From large triangles take away the smaller triangles, and the remainders, the complements, are equal (ax. 3).

(N.B. This method of proof is very frequently adopted in Book II.)

I. 44. A somewhat complicated problem follows: To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. By the phrase 'apply a parallelogram to a given straight line' is meant that the given line shall form one side of the parallelogram to be applied; the problem might have been enunciated as I. 1 and I. 46; on a given straight line to describe, &c.

You will observe that in the parallelogram required we have fixed the area, one pair of sides, and one pair of angles (since we know that the opposite sides and angles of a parallelogram are equal, I. 33). We can by I. 42 describe a parallelogram fulfilling two of these conditions, i.e. equal in area to the given triangle and having an angle equal to the given angle. Draw this parallelogram, calling it EG. The problem now consists in describing a parallelogram on AB equal in area to EG, and also having an angle equal to one of the angles of EG. We must ask under what conditions are parallelograms equal in area? Propositions XXXV. and XXXVI. tell us, when they are on equal or the same bases and between the same parallels, but a moment's consideration I will show that we cannot obtain these conditions here. Have we had any other case of equality of area in parallelograms?

If

A

B

K

H

Ꮹ

D

F

you will refer to the last proposition you will see that the 'complements' which we proved equal are parallelograms, and that none of the sides of one complement is of necessity equal to one of those of the other, neither are they E between the same parallels. And farther, if we examine the figure of I. 43, we shall find that it meets all the conditions necessary for the solution of the problem. Let EK represent the line to which the parallelogram is to be applied, and let HF represent the parallelogram equal to the given triangle, and having an angle HKF equal to the given angle. Then EG will be the parallelogram required. For it is applied to line EK, it is equal to HF, the other complement, and it has an angle EKG equal to HKF, since they are opposite vertical angles (I. 15).

G

F

Can we retrace our steps and construct a similar figure? Let us try. Let AB be the given line, C the given triangle, and D the given angle. By I. 42 we can, as we have already remarked, construct a parallelogram EG (which will represent HF in the figure above), i.e. equal in area to C, and having an angle at G equal to the given angle D, and we must take care to place EG in such a position that one of its sides shall be in a straight line with AB, and that the angle BGF shall correspond with the angle HKF, thusThen AB, BG will be adjacent sides of one of the parallelograms' about the diameter,' and we can complete it by drawing through A a line parallel to BG, and producing FG to meet it in H. The diameter will be formed by joining HB, and this diameter produced will be the diameter of the large parallelogram, consequently it must meet the opposite side of the large parallelogram which we can obtain by producing FE. But are we sure that HB and FE produced will meet? We know that any

H

G

B

A

B

E

E

F

straight lines which with another straight line make the two

Η

G

F

interior angles on the same side together less than two right angles, if produced, will meet on that side (def. 35). We have, then, to ask, are the angles BHF, HFE together less than two right angles? It is evident that the angle BHF is less than the angle AHF of which it is a part, and since AH and FE are parallel, the angles AHF, HFE are equal to two right angles; therefore, the angles BHF, HFE are together less than two right angles, and, consequently, HB and FE will meet if produced towards BE. Let them be produced and meet in K. The triangle HFK thus formed will be half of the large parallelogram which we are trying to construct, and which will be completed by drawing a line through K parallel to HF, and producing HA to meet it in L. To obtain the other complement, which is the required parallelogram applied to AB, we have only to produce GB to meet LK in M.

L

B

E

M

SUMMARY OF PROPOSITION XLIV., PROBLEM 12.

To a given straight line to apply a parallelogram which shall be equal to a given triangle, and have one of its angles equal to a given angle.

Cons.-Make gram BGFE equal to given triangle, and having EBG equal to given angle. Place it so that BE, one of the arms of EBG, is in the same straight line with

AB (I. 42). (One com-
Through A

plement.)

about diameter.) HB and FE will meet if produced, since angles BHF, HFE are less than two right angles. Let them meet in K. Through K draw a line parallel to HF, and produce HA to meet it in L. (Large parallelogram.) Produce GB to meet KL in M. (Remaining |gram and complement.)

Proof.-Comps. BL and BF of parallelogram HLKF are

Therefore to AB a gram BL has been applied, equal to ▲ C, and having an angle ABM equal to ▲ D.—Q.E.F.

I. 45. The next problem is to describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given angle. In Proposition XXXII. we learnt how to divide any rectilineal figure into triangles by drawing lines from one of its angles to all of the opposite angles. The figure given in the pro- A blem has four sides, and therefore can be divided into two triangles by joining either pair of opposite angles. Let ABCD be the given rectilineal figure thus divided by BD, and let E be the given angle.

B

Now by I. 42 we can construct a parallelogram equal to one of these triangles, and by the last proposition we can apply to one side of this parallelogram another, equal to the remaining triangle, and in each case we can make one of the angles equal to the given angle E. And since we wish the whole figure to form a parallelogram we must take the angle equal to E in the second parallelogram opposite to the corresponding angle in the first, thus: Let FGHK be the parallelogram constructed equal to triangle ABD, and having its angle FKH equal to angle E (I. 42), and to its side GH apply a parallelogram GLMH equal to triangle DCB, and having angle GLM equal to angle E.

Then FKML is a figure consisting of two parallelograms which are together equal to the given figure ABCD, and it has an angle equal to the given angle E. It remains to show that it is a parallelogram, i.e. that its opposite sides are parallel.

I. The sides FK and LM are each of them parallel to GH

(a common side to the two parallelograms). Therefore, they are parallel to each other.

F

G

L

Again, FG, GL are parallel to KH, HM, each to each. But are these parts in the same straight line so that they form a side of the parallelogram? What test have we had for showing when two straight lines are 'in one and the same straight line'? We must refer to Proposition XIV., which tells us that 'when a straight line meeting two other straight lines makes the adjacent angles equal to two right angles, these straight lines are in one and the same straight line.'

K

II

M

(1) Are angles GHK, GHM equal to two right angles? If so KHM is a straight line.

(2) Also, are angles FGH, LGH equal to two right angles? If so FGL is a straight line.

=

Then, examining the angles, we know that angles FKH, GHK are equal to two right angles, since GH is parallel to FK (I. 29), and angle GHM = angle GLM angle FKH. Therefore, the angles GHK, GHM are together equal to two right angles, and KHM is a straight line. Similarly angles HGL, GLM are equal to two right angles (I. 29). But angle FGH = ▲ FKH = / GLM (I. 33). Therefore, angles FGH, HGL = two right angles, and FGL is a straight line, and it is parallel to KHM. It follows that the figure FKML is a parallelogram, and we have shown that it is equal to ABCD, and that it has an angle FKM equal to E.

In the same way a parallelogram may be constructed equal to a rectilineal figure of any number of sides, hy resolving the figure into triangles, then by I. 42 drawing a parallelogram equal to one of the triangles, and applying successively parallelograms equal to the other triangles.

NOTE.-In showing that FGL is a straight line Euclid reasons thus: The angles MHG, HGL are equal to two right angles (I. 29), ard the alternate angles FGH, GHM are equal; therefore the angles FGH, LGH are equal to two right angles, and FGL is a straight line, and it is parallel to KHM.