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PROPOSITION XLII. 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 ▲, and D the given 4. It is required to describe a ☐ equal to ▲ ABC, having one of its 48= L D.

Bisect BC in E and join AE.

At E make 4 CEF= L D.

Draw AFG || to BC, and from C draw CG || to EF.
Then FECG is a parallelogram.

Now Δ AEB= Δ AEC,

they are on equal bases and between the same ||s;

:. Δ ABC is double of Δ AEC.

I. 38.

But FECG is double of ▲ AEC,

they are on same base and between same s;
:. □ FECG = ▲ ABC,

I. 41.

Ax. 6.

and FECG has one of its 4s, CEF= LD. :.FECG has been described as was reqd.

Q. E. F.

Ex. 1. Describe a triangle, which shall be equal to a given parallelogram, and have one of its angles equal to a given rectilineal angle.

Ex. 2. Construct a parallelogram, equal to a given triangle, and such that the sum of its sides shall be equal to the sum of the sides of the triangle.

Ex. 3. The perimeter of an isosceles triangle is greater than the perimeter of a rectangle, which is of the same altitude with, and equal to, the given triangle.

PROPOSITION XLIII. THEOREM.

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

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Let ABCD be a, of which BD is a diagonal, and EG, HK the □s about BD, that is, through which BD passes,

and AF, FC the others, which make up the whole figure ABCD,

and which are .. called the Complements.

Then must complement AF=complement FC.

For BD is a diagonal of □ AC,

:. ▲ ABD= ^ CDB;

I. 34.

and BF is a diagonal of □ HK,

:. ▲ HBF= ▲ KFB;

and FD is a diagonal of EG,

.. A EFD= ▲ GDF.

Hence sum of As HBF, EFD= sum of As KFB, GDF.
Take these equals from As ABD, CDB respectively,
AF remaining FC.

then remaining

Ax. 3.
Q. E. D.

Ex. 1. If through a point 0, within a parallelogram ABCD, two straight lines are drawn parallel to the sides, and the parallelograms OB, OD are equal; the point is in the diagonal AC.

Ex. 2. ABCD is a parallelogram, AMN a straight line meeting the sides BC, CD (one of them being produced) in M, N. Shew that the triangle MBN is equal to the triangle MDC.

PROPOSITION XLIV. PROBLEM.

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 st. line, C the given ▲, D the

given 4.

It is required to apply to AB a ☐ of its 48=

LD.

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I. 42.

Make a =AC, and having one of its angles = ▲ D,

and suppose it to be removed to such a position that one of the sides containing this angle is in the same st. line with AB, and let the be denoted by BEFG.

Produce FG to H, draw AH || to BG or EF, and join BH.
Then FH meets the 8 AH, EF,

.. sum of 48 AHF, HFE=two rt. 48;

.. sum of 48 BHG, HFE is less than two rt. 28;

.. HB, FE will meet if produced towards B, E.

Let them meet in K.

Through K draw KL || to EA or FH,

Post. 6.

and produce HA, GB to meet KL in the pts. L, M. Then HFKL is a □, and HK is its diagonal;

and AG, ME ares about HK,

.. complement BL= complement BF,

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Also the BL has one of its 4s, ABM = LEBG, and

.. equal to 4 D.

Q. E. F.

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To describe a parallelogram, which shall be equal to a given rectilinear figure, and have one of its angles equal to a given angle.

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Let ABCD be the given rectil. figure, and E the given 4.
Join AC.

It is required to describe a = to ABCD, having one of its LS LE.

Describe a FGHK= ▲ ABC, having FKH= L E.

To GH apply a □ GHML= ▲ CDA, having GHM:

I. 42.

=LE.

Then FKML is the

reqd.

For LGHM and

FKH are each = LE;

I. 44.

.. LGHM = L FKH,

.. sum of 48 GHM, GHK=sum of 48 FKH, GHK

=two rt. 48;

.. KHM is a st. line.

I. 29.

I. 14.

Again, HG meets the Is FG, KM,

LFGH= LGHM,

.. sum of 48 FGH,LGH=sum of 48 GHM, LGH

=two rt. 48;

.. FGL is a st. line.

Then KF is || to HG, and HG is || to LM,

.. KF is to LM;

and KM has been shewn to be || to FL,

.. FKML is a parallelogram,

and FHA ABC, and GM= ▲ CDA,

and

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I. 20.

I. 14.

I. 30.

In the same way a may be constructed equal to a given rectil. fig. of any number of sides, and having one of its angles equal to a given angle.

Q. E. F.

Miscellaneous Exercises.

1. If one diagonal of a quadrilateral bisect the other, it divides the quadrilateral into two equal triangles.

2. If from any point in the diagonal, or the diagonal produced, of a parallelogram, straight lines be drawn to the opposite angles, they will cut off equal triangles.

3. In a trapezium the straight line, joining the middle points of the parallel sides, bisects the trapezium.

4. The diagonals AC, BD of a parallelogram intersect in O, and P is a point within the triangle AOB; prove that the difference of the triangles APB, CPD is equal to the sum of the triangles APC, BPD.

5. If either diameter of a parallelogram be equal to a side of the figure, the other diameter shall be greater than any side of the figure.

6. If through the angles of a parallelogram four straight lines be drawn parallel to its diagonals, another parallelogram will be formed, the area of which will be double that of the original parallelogram.

7. If two triangles have two sides respectively equal and the included angles supplemental, the triangles are equal.

8. Bisect a given triangle by a straight line drawn from a given point in one of the sides.

9. If the base of a triangle ABC be produced to a point D such that BD is equal to AB, and if straight lines be drawn from A and D to E, the middle point of BC; prove that the triangle ADE is equal to the triangle ABC.

10. Prove that a pair of the diagonals of the parallelograms, which are about the diameter of any parallelogram, are parallel to each other.

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