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PROPOSITION 41.

THEOREM. If a parallelogram and a triangle be on the same base and between the same parallels, the parallelogram shall be double of the triangle.

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Let the parallelogram ABCD, and the triangle EBC be upon the same base BC, and between the same parallels BC, AE. Then shall the parallelogram ABCD be double of the triangle EBC. Construction,

Join AC. Proof. Now the triangles ABC, EBC are on the same base BC, and between the same parallels BC, AE ; .. the triangle ABC = the triangle EBC.

I. 37. And since the diagonal AC bisects ABCD;

I. 34. .. the parallelogram ABCD is double of the triangle ABC.

Therefore the parallelogram ABCD is also double of the triangle EBC.

Q.E.D.

EXERCISES.

1. ABCD is a parallelogram, and X, Y are the middle points of the sides AD, BC ; if Z is any point in XY, or XY produced, shew that the triangle AZB is one quarter of the parallelogram ABCD.

2. Describe a right-angled isosceles triangle equal to a given square.

3. If ABCD is a parallelogram, and X, Y any points in DC and AD respectively : shew that the triangles AXB, BYC are equal in area.

4. ABCD is a parallelogram, and P is any point within it ; shew that the sum of the triangles PAB, PCD is equal to half the parallelogram.

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.

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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. Construction. Bisect BC at E.

I. 10. At E in CE, make the angle CEF equal to D; I. 23.

through A draw AFG parallel to EC; I. 31. and through c draw CG parallel to EF. Then FECG shall be the parallelogram required.

Join AE. Proof. 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, being on the same base EC, and between the same parallels

EC and AG. Therefore the parallelogram FECG is equal to the triangle and it has one of its angles CEF equal to the given angle D.

Q.E.F.

1. 41,

ABC;

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 stand. ing on the same base. When does the construction fail ?

8.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 Paralleiograms 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.

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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.

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I. 34.

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. 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,

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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.

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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 1. 42*. Through A draw AH parallel to BG or EF, to meet FG produced in H.

1. 31. Join HB. Then because AH and EF are parallel, and HF meets them, is 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 HĂ, GB to meet KL in the points L and M.

Then shall BL be the parallelogram required.

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