PROPOSITION 42. PROBLEM.

To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given angle.

Let ABC be the given triangle, and D the given angle. It is required to describe a parallelogram equal to ABC, and having one of its angles equal to D.

At E in CE, make the angle CEF equal to D;
through A draw AFG parallel to EC;
and through C draw CG parallel to EF.
Then FECG shall be the parallelogram required.
Join AE.

Proof.

I. 10.

I. 23.

I. 31.

Now the triangles ABE, AEC are on equal bases

BE, EC, and between the same parallels;

=

.. the triangle ABE the triangle AEC; I. 38. .. the triangle ABC is double of the triangle AEC.

But FECG is a parallelogram by construction; Def. 36. and it is double of the triangle AEC,

I. 41.

being on the same base EC, and between the same parallels
EC and AG.
Therefore the parallelogram FECG is equal to the triangle
ABC ;

and it has one of its angles CEF equal to the given angle D.

Q.E.F.

EXERCISES.

1. Describe a parallelogram equal to a given square standing on the same base, and having an angle equal to half a right angle.

2. Describe a rhombus equal to a given parallelogram and standing on the same base. When does the construction fail?

H.S.E.

DEFINITION. If in the diagonal of a parallelogram any point is taken, and straight lines are drawn through it parallel to the sides of the parallelogram; then of the four parallelograms into which the whole figure is divided, the two through which the diagonal passes are called Parallelograms about that diagonal, and the other two, which with these make up the whole figure, are called the complements of the parallelograms about the diagonal.

Thus in the figure given above, AEKH, KGCF are parallelograms about the diagonal AC; and the shaded figures HKFD, EBGK are the complements of those parallelograms.

NOTE. A parallelogram is often named by two letters only, these being placed at opposite angular points.

PROPOSITION 43. THEOREM.

The complements of the parallelograms about the diagonal of any parallelogram, are equal to one another.

Let ABCD be a parallelogram, and KD, KB the complements of the parallelograms EH, GF about the diagonal AC. Then shall the complement BK be equal to the complement KD.

Proof.

I. 34.

Because EH is a parallelogram, and AK its diagonal, .. the triangle AEK the triangle AHK. Similarly the triangle KGC=the triangle KFC. Hence the triangles AEK, KGC are together equal to the triangles AHK, KFC.

But since the diagonal AC bisects the parallelogram ABCD; .. the whole triangle ABC= the whole triangle ADC. I. 34. Therefore the remainder, the complement BK, is equal to the remainder, the complement KD.

Q.E.D.

EXERCISES.

In the figure of Prop. 43, prove that

(i) The parallelogram ED is equal to the parallelogram BH.
(ii) If KB, KD are joined, the triangle AKB is equal to the
triangle AKD.

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.

Let AB be the given straight line, C the given triangle, and D the given angle.

It is required to apply to the straight line AB a parallelogram equal to the triangle C, and having an angle equal to the angle D.

Construction. On AB produced describe a parallelogram BEFG equal to the triangle C, and having the angle EBG equal to the angle D. I. 22 and I. 42*. Through A draw AH parallel to BG or EF, to meet FG produced in H.

Join HB.

I. 31.

Then because AH and EF are parallel, and HF meets them, ... the angles AHF, HFE together = two right angles. I. 29. Hence the angles BHF, HFE are together less than two right angles;

.. HB and FE will meet if produced towards B and E. Ax. 12. Produce HB and FE to meet at K.

Through K draw KL parallel to EA or FH; I. 31. and produce HA, GB to meet KL in the points L and M. Then shall BL be the parallelogram required.

Proof. Now FHLK is a parallelogram,

Constr.

and LB, BF are the complements of the parallelograms about the diagonal HK:

.. the complement LB the complement BF. I. 43.

=

But the triangle C-the figure BF;

... the figure LB = the triangle C.

Constr.

Again the angle ABM = the vertically opposite angle GBE; also the angle D = the angle GBE;

.. the angle ABM = the angle D.

Constr.

Therefore the parallelogram LB, which is applied to the straight line AB, is equal to the triangle C, and has the angle ABM equal to the angle D.

Q.E.F.

This step of the construction is effected by first describing on AB produced a triangle whose sides are respectively equal to those of the triangle C (1. 22); and by then making a parallelogram equal to the triangle so drawn, and having an angle equal to D (1. 42).

1.

QUESTIONS FOR REVISION.

Quote Euclid's Twelfth Axiom. What objections have been raised to it, and what substitute for it has been suggested?

2. Which of Euclid's Propositions, dealing with parallel straight lines, depends on Axiom 12? Furnish an alternative proof.

3. Straight lines which are parallel to the same straight line are parallel to one another [Prop. 30]. Deduce this from Playfair's

Axiom.

4. Define a parallelogram, an altitude of a triangle, a median of a triangle, parallelograms about the diagonal of a parallelogram.

5. What is meant by superposition? On what Axiom does this method depend? Give instances of figures which are equal in area, but which cannot be superposed.

6. In fig. 2 of Prop. 35 shew how one parallelogram may be cut into pieces, which, when fitted together in other positions, make up the other parallelogram.